integer values of polynomials

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Integer values of polynomialsGiulio Peruginelli

To cite this version:Giulio Peruginelli. Integer values of polynomials. Number Theory [math.NT]. Università degli studidi Pisa, 2008. English. tel-00796349

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Contents

Introduction 3

1 Algebraic function fields in one variable 91.1 Luroth theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2 Minimal couples of rational functions . . . . . . . . . . . . . . . . . . . . . 111.3 Valuation rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Ritt’s decomposition theorem for polynomials 172.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Monodromy and Galois covering . . . . . . . . . . . . . . . . . . . . . . . 21

2.2.1 Ramified coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2.2 Galois coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.3 Blocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.4 First theorem of Ritt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3 Plane algebraic curves 433.1 Affine curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.2 Projective curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.3 Rational maps between curves . . . . . . . . . . . . . . . . . . . . . . . . . 473.4 Geometric points and places . . . . . . . . . . . . . . . . . . . . . . . . . . 503.5 Parametrization with rational coefficients . . . . . . . . . . . . . . . . . . 513.6 Standard parametrization of rational curves . . . . . . . . . . . . . . . . . 523.7 Polynomial parametrization of rational curves . . . . . . . . . . . . . . . . 54

4 Curves of Schinzel 574.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.2 Symmetric case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.3 Kubota’s results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5 Parametrization of integer-valued polynomials 655.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655.2 Linear factors of bivariate separated polynomials . . . . . . . . . . . . . . 675.3 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.4 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

1

2 CONTENTS

5.4.1 Case f(Z) = g(Z), with g ∈ Z[X] . . . . . . . . . . . . . . . . . . . 775.4.2 General case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.5 Number field case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Bibliography 88

CONTENTS 3

Introduction

Let f(X) be a polynomial with rational coefficients, S be an infinite subset of therational numbers and consider the image set f(S). If g(X) is a polynomial such thatf(S) = g(S) we say that g parametrizes the set f(S). Besides the obvious solution g = fwe may want to impose some conditions on the polynomial g; for example, if f(S) ⊂ Zwe wonder if there exists a polynomial with integer coefficients which parametrizes theset f(S).

Moreover, if the image set f(S) is parametrized by a polynomial g, there comes thequestion whether there are any relations between the two polynomials f and g. Forexample, if h is a linear polynomial and if we set g = f h, the polynomial g obviouslyparametrizes the set f(Q). Conversely, if we have f(Q) = g(Q) (or even f(Z) = g(Z))then by Hilbert’s irreducibility theorem there exists a linear polynomial h such thatg = f h. Therefore, given a polynomial g which parametrizes a set f(S), for an infinitesubset S of the rational numbers, we wonder if there exists a polynomial h such thatf = g h. Some theorems by Kubota give a positive answer under certain conditions.

The aim of this thesis is the study of some aspects of these two problems related tothe parametrization of image sets of polynomials.

In the context of the first problem of parametrization we consider the followingsituation: let f be a polynomial with rational coefficients such that it assumes integervalues over the integers. Does there exist a polynomial g with integer coefficients suchthat it has the same integer values of f over the integers?

This kind of polynomials f are called integer-valued polynomials. We remark that theset of integer-valued polynomials strictly contains polynomials with integer coefficients:take for example the polynomial X(X − 1)/2, which is integer-valued over the set ofintegers but it has no integer coefficients. So, if f is an integer-valued polynomial,we investigate whether the set f(Z) can be parametrized by a polynomial with integercoefficients; more in general we look for a polynomial g ∈ Z[X1, . . . , Xm], for somenatural number m ∈ N, such that f(Z) = g(Zm). In this case we say that f(Z) isZ-parametrizable.

In a paper of Frisch and Vaserstein it is proved that the subset of pythagoreantriples of Z3 is parametrizable by a single triple of integer-valued polynomials in fourvariables but it cannot be parametrized by a single triple of integer coefficient poly-nomials in any number of variables. In our work we show that there are examples ofsubset of Z parametrized by an integer-valued polynomial in one variable which cannotbe parametrized by an integer coefficient polynomial in any number of variables.

If f(X) is an integer-valued polynomial, we give the following characterization ofthe parametrization of the set f(Z): without loss of generality we may suppose thatf(X) has the form F (X)/N , where F (X) is a polynomial with integer coefficients andN is a minimal positive integer. If there exists a prime p different from 2 such that pdivides N then f(Z) is not Z-parametrizable. If N = 2n and f(Z) is Z-parametrizablethen there exists a rational number β which is the ratio of two odd integers such thatf(X) = f(−X + β). Moreover f(Z) = g(Z) for some g ∈ Z[X] if and only if f ∈ Z[X]

4 CONTENTS

or there exists an odd integer b such that f ∈ Z[X(b−X)/2]. We show that there existsinteger-valued polynomials f(X) such that f(Z) is Z-parametrizable with a polynomialG(X1, X2) ∈ Z[X1, X2], but f(Z) 6= g(Z) for every g ∈ Z[X].

In 1963 Schinzel gave the following conjecture: let f(X,Y ) be an irreducible poly-nomial with rational coefficients and let S be an infinite subset of Q with the propertythat for each x in S there exists y in S such that f(x, y) = 0; then either f is linear inY or f is symmetric in the variables X and Y .

We remark that if a curve is defined by a polynomial with Schinzel’s property thenits genus is zero or one, since it contains infinite rational points; here we use a theorem ofFaltings which solved the Mordell-Weil conjecture (if a curve has genus greater or equalto two then the set of its rational points is finite). We will focus our attention on thecase of rational curves (genus zero). Our objective is to describe polynomials f(X,Y )with Schinzel’s property whose curve is rational and we give a conjecture which saysthat these rational curves have a parametrization of the form (ϕ(T ), ϕ(r(T ))).

This problem is related to the main topic of parametrization of image sets of polyno-mials in the following way: if (ϕ(T ), ψ(T )) is a parametrization of a curve f(X,Y ) = 0(which means f(ϕ(T ), ψ(T )) = 0), where f is a polynomial with Schinzel’s property, letS = ϕ(t)|t ∈ S′ be the set of the definition of Schinzel, where S′ ⊂ Q. Then for eacht ∈ S′ there exists t′ ∈ S′ such that ψ(t) = ϕ(t′), hence ψ(S′) ⊂ ϕ(S′). So, in the caseof rational curves, the problem of Schinzel is related to the problem of parametrizationof rational values of rational functions with other rational functions (we will show thatunder an additional hypothesis we can assume that (ϕ(T ), ψ(T )) are polynomials). Inparticular, if (ϕ(T ), ψ(T )) is a parametrization of a curve defined by a symmetric poly-nomial, then ψ(T ) = ϕ(a(T )), where a(T ) is an involution (that is a a = Id). So inthe case of rational symmetric plane curves we have this classification in terms of theparametrization of the curve.

It turns out that this argument is also related to Ritt’s theory of decomposition ofpolynomials. His work is a sort of ”factorization” of polynomials in terms of indecompos-able polynomials, that is non-linear polynomials f such that there are no g, h of degreeless than deg(f) such that f = g h. The indecomposable polynomials are some sort of”irreducible” elements of this kind of factorization.

In the first chapter we recall some basic facts about algebraic function fields in onevariable, the algebraic counterpart of algebraic curves. In particular we state the famousLuroth’s theorem, which says that a non trivial subextension of a purely trascendentalfield of degree one is purely trascendental.

We give the definition of minimal couple of rational functions that we will use later tocharacterize algebraically a proper parametrization of a rational curve. We conclude thechapter with the general notion of valuation ring of a field and we characterize valuationrings of a purely trascendental field in one variable (which corresponds geometrically to

CONTENTS 5

the Riemann sphere, if for example the base field is the field of the complex numbers).Moreover valuation rings of algebraic function fields in one variable are discrete valuationrings.

In the second chapter we state the first theorem of Ritt, which deals with decompo-sition of polynomials with complex coefficients with respect to the operation of compo-sition. In a paper of 1922 Ritt proved out that two maximal decompositions (that is adecomposition whose components are neither linear nor further decomposable) of a com-plex polynomial have the same number of components and their degrees are the sameup to the order. We give a proof in the spirit of the original paper of Ritt, which usesconcepts like monodromy groups of rational functions, coverings and theory of blocks inthe action of a group on a set.

This result can be applied in the case of an equation involving compositions of poly-nomials: thanks to Ritt’s theorem we know that every side of the equation has the samenumber of indecomposable component.

In the third chapter we give the classical definition of plane algebraic curves, bothin the affine and projective case. We show that there is a bijection between the pointsof a non-singular curve and the valuation rings of its rational function field (which arecalled places of the curve). More generally speaking, if we have a singular curve C, theset of valuation rings of its rational function field is in bijection with the set of points ofa non-singular model C ′ of the curve (that is the two curves C and C ′ are birational),called desingularization of the curve.

Then we deal with curves whose points are parametrized by a couple of rationalfunctions in one parameter; we call these curves rational. From a geometric point ofview a rational curve has desingularization which is a compact Riemann surface of genuszero, thus isomorphic to P1. Finally we expose some properties of parametrizations ofrational curves; we show a simple criterium which provides a necessary and sufficientcondition that lets a rational curve have a polynomial parametrization in terms of placesat infinity.

In the fourth chapter we study the aforementioned conjecture of Schinzel.For example, if f(X,Y ) = Y −a(X) then by taking S the full set of rational numbers

we see that the couple (f, S) satisfies the Schinzel’s property. If f is symmetric and theset of rational points of the curve determined by f is infinite, then if we define S tobe the projection on the first coordinate of the rational points of the curve we obtainanother example of polynomial with the above property.

The hypothesis of irreducibility of the polynomial f is required because we want toavoid phenomenon such as f(X,Y ) = X2−Y 2 and S = Q, where f is neither linear norsymmetric. In general if a polynomial f(X,Y ) has X − Y as a factor, then it admitsthe full set of rational numbers as set S. Another example is the following (privatecommunication of Schinzel): let

f(X,Y ) = (Y 2 −XY −X2 − 1)(Y 2 −XY −X2 + 1)

6 CONTENTS

and S = Fnn∈N, where Fn is the Fibonacci sequence which satisfies the identityF 2n+1−Fn+1Fn−F

2n = (−1)n for each natural number n; if f1, f2 ∈ Q[X,Y ] are the two

irreducible factors of f then for each n ∈ N the couple of integers (Fn, Fn+1) is a pointof the curve associated to the polynomial f1 or f2, according to the parity of n.

Zannier has recently given the following counterexample to Schinzel’s conjecture:

f(X,Y ) = Y 2 − 2(X2 +X)Y + (X2 −X)2

with S equal to the set of rational (or integer) squares. The idea is the following:it is well known that for each couple of rational functions (ϕ(t), ψ(t)) with coefficientsin a field k there exists a polynomial f ∈ k[X,Y ] such that f(ϕ(t), ψ(t)) = 0. In factk(t) has trascendental degree one over k; we also say that ϕ and ψ are algebraicallydependent. Moreover if we require that the polynomial f is irreducible then it is uniqueup to multiplication by constant.

This procedure allows us to build families of polynomials with Schinzel’s property:it is sufficient to take couples of rational functions (ϕ(t), ϕ(r(t))), where ϕ(t), r(t) arerational functions. If we consider the irreducible polynomial f ∈ Q[X,Y ] such thatf(ϕ(t), ϕ(r(t))) = 0 and the set S = ϕ(t)|t ∈ Q, we see that (f, S) has Schinzel’s prop-erty. In particular Zannier’s example is obtained from the couple of rational functions(ϕ(t), r(t)) = (t2, t(t + 1)). If deg(ϕ) > 1 and deg(r(t)) > 1 then it turns out that f isneither linear nor symmetric in X and Y , but it is a polynomial with Schinzel’s property.

In the last chapter we deal with the problem of parametrization of integer-valuedpolynomials and we prove the results mentioned at the beginning of this introduction.The idea of the proof is the following: let f(X) = F (X)/N be an integer-valued poly-nomial as above; since the set of integer-valued polynomials is a module over Z, we canassume that N is a prime number p. We remark that a bivariate polynomial of the formf(X) − f(Y ) has over Q only two linear factors; moreover, the set of integer values nsuch that there exists q ∈ Q such that (n, q) belongs to an irreducible component of thecurve f(X)−f(Y ) = 0 which is not linear in Y , has zero density, by a theorem of Siegel.If f(Z) is Z-parametrizable by a polynomial g ∈ Z[X1, . . . , Xm] = Z[X] then by Hilbert’sirreducibility theorem there exists Q ∈ Q[X] such that F (Q(X)) = pg(X); we obtainnecessary conditions for such polynomial Q in order to satisfy the previous equality. Inthe same hypothesis, for each n ∈ Z there exists xn ∈ Zm such that f(n) = f(Q(xn)).So we study how the points (n,Q(xn)), for n ∈ Z, distribute among the irreducible com-ponents of the curve f(X)− f(Y ) = 0; by the aforementioned theorem of Siegel it turnsout that, up to a subset of density zero of Z, they belong to components determined bylinear factors of f(X)− f(Y ). For each of them, the projection on the first componentof this kind of points is a set of integers contained in a single residue class modulo theprime p. So if p is greater then two, which is the maximum number of linear factors ofa bivariate separated polynomial over Q, the set f(Z) is not Z-parametrizable.

The problem of factorization of bivariate separated polynomials, that is polynomialsof the form f(X) − g(Y ), is a topic which has been intensively studied for years (Bilu,Tichy, Zannier, Avanzi, Cassou-Nogues, Schinzel, etc...)

CONTENTS 7

Our next aim is the classification of the integer-valued polynomials f(X) such thatf(Z) is parametrizable with an integer coefficient polynomial in more than one variable(for example f(X) = 3X(3X − 1)/2). I conjecture that such polynomials (except whenf ∈ Z[X]) belong to Z[pkX(pkX−a)/2], where p is a prime different from 2, a is an oddinteger coprime with p and k a positive integer. I show in my work that if f(X) is sucha polynomial, then f(Z) is Z-parametrizable.

Moreover we want to study the case of number fields, that is the parametrization ofsets f(OK), where OK is the ring of integers of a number field K and f ∈ K[X] suchthat f(OK) ⊂ OK , with polynomials with coefficients in the ring OK .

Acknowledgements I wish to warmly thank prof. Umberto Zannier forsuggesting me the problem of parametrization of integer-valued polynomials and theuseful help he gave me in the proof of the theorem which describes these parametriza-tions. Without his help I couldn’t have stated and proved this theorem in such a niceand simple form. He also pointed me out two mistakes I made.

I am also in debt with prof. Sheram S. Abhyankar and prof. Andrzej Schinzel. I hadwith them useful discussions which allowed me to see some problems in greater depth.In particular I owe Schnizel a debt of gratitude for the time he dedicated to me listeningto the problems arosen in the development of my work of thesis.

I want to thank also prof. Mario Poletti who read some pages of my work and pointedme out some errors.

I wish to thank also all my colleagues in Pisa, especially Lorenzo Brasco, LucaCaputo, Sara Checcoli and Laura Paladino. They helped me both with mathematicsand the English language. For the latter I want to thank my brother’s wife Elena, whosuffered in reading something she couldn’t understand the meaning of.

8 CONTENTS

Chapter 1

Algebraic function fields in one

variable

The algebraic analogue of algebraic plane curves, that is the zero-locus of a polyno-mial f(X,Y ) (see chapter 4), are algebraic function fields in one variable; we are goingto explain some basic facts about them.

Let k be a field which from now on we call the base field; an algebraic functionfield in one variable over k is a field K finitely generated over k such that thetrascendence degree of K over k is equal to 1 and k is algebraically closed in K.

1.1 Luroth theorem

The ”simplest”, so to speak, algebraic function field in one variable is the puretrascendental field in one indeterminate k(t). Here we give some details about this kindof algebraic function fields.

Definition 1.1.1 If f(t) = ϕ(t)ψ(t) is a rational function in reduced form (that is ϕ and ψ

are coprime polynomials) then the degree of f is defined as

deg(f(t)) + maxdeg(ϕ(t)),deg(ψ(t))

Lemma 1.1.2 If f(t) is a rational function over k then the extension of field k(f) ⊂ k(t)is finite and its degree is equal to deg(f).

Proof : Let f(t) be of the form r(t)s(t) where r, s ∈ k[t] are coprime polynomials; then

t satisfies the following polynomial with coefficients in K(f):

F (X) = r(X)− f · s(X)

9

10 CHAPTER 1. ALGEBRAIC FUNCTION FIELDS IN ONE VARIABLE

This polynomial is irreducible in k[f,X] = k[X][f ] because it has degree 1 in f andit is primitive over the ring k[X]; by Gauss Lemma it is irreducible in k(f)[X]. TheX-degree of F is equal to deg(f).

Every non-trivial subextension of a pure trascendental extension of degree one is apure trascendental extension of degree one, as the following theorem says (this is falsefor trascendental extension of higher degree).

Theorem 1.1.3 (Luroth) Let k be a field and t a trascendental element over k. If Kis a field such that k ( K ⊂ k(t) then there exists a rational function g in k(t) such thatK = k(g).

Proof : The field k(t) is a finite algebraic extension of K because if g ∈ K − k thenk(t)/k(g) is a finite extension by previous lemma.

Let

f(X) = Xn + an−1(t)Xn−1 + . . .+ a0(t)

be the minimal polynomial of t over K, where ai(t) ∈ K ⊂ k(t) for each i = 0, . . . ,n− 1; since t is trascendental over k, there exists an index j ∈ 0, . . . , n− 1 such thataj(t) 6∈ k. If we multiply f(X) by the least common multiple of the denominators of thecoefficients ai(t)’s we obtain a primitive irreducible polynomial in k[t,X]

f0(t,X) = αn(t)Xn + . . .+ α0(t)

where αi ∈ k[t] for each i = 0, . . . , n ; let m be the degree of f0(t,X) in t, that is themaximum of the degrees of the αi(t)’s.

If θ = p(t)/q(t) is the reduced representation of aj(t), where p, q ∈ k[t] are coprime,then t is root of the polynomial

H(X) = q(X)θ − p(X)

with coefficients in the field k(θ) ⊂ K; indeed H(X) is the minimal polynomialof t over k(θ). So f(X) divides H(X) in K[X] and consequently f0(t,X) divides theprimitive polynomial H0(t,X) = q(X)p(t)− p(X)q(t) in k[t,X]. We have the followingequality in k[t,X]

q(X)p(t)− p(X)q(t) = f0(t,X)s(t,X)

where s ∈ k[t,X].Observe now that the degree in t of the first member is less or equal than m, since θ

is a coefficient of f(X); then the degree in t of s(t,X) is equal to 0. So s(t,X) = s(X) isa polynomial in the variable X and it divides H0(t,X) which is primitive in k[X]: thisimplies that the polynomial s is constant. Hence the degree in X of H0(t,X), which isthe degree [k(t) : k(θ)], is equal to the degree in X of f(t,X). Hence we have provedthat K = k(θ).

1.2. MINIMAL COUPLES OF RATIONAL FUNCTIONS 11

The proof works for each non-constant coefficients ai(t) of f(X): the field K can begenerated over k by each coefficient ai(t) which is not constant.

The following theorem is the polynomial version of the Luroth theorem: if a non-trivial subextension of k(t) contains a polynomial then it can be generated by a polyno-mial.

Theorem 1.1.4 (Luroth in polynomial form) Let k be a field and t a trascendentalelement over k. If K is a field such that k ( K ⊂ k(t) and it contains a polynomialf ∈ K[t], then there exists g ∈ k[t] such that K = k(g).

A proof of this theorem was already known to Ritt (see [35]) and it is the one we aregoing to show.

Proof : From Luroth’s theorem there exists a rational function r ∈ k(t) such thatK = k(r); from the inclusion of fields k(f) ⊂ k(r) ⊂ k(t) it follows that f = s r wheres ∈ K(r).

The natural map

F : k ∪ ∞ → k ∪ ∞

t 7→ f(t)

associated to the polynomial f is totally ramified over ∞, which means that the fiber ofF over ∞ has only one point since F−1(∞) = ∞; we obviously have that F = S R,where S and R are the maps from k ∪ ∞ in itself associated to s and r respectively.

This fact implies that the maps S and R are totally ramified over ∞ and R(∞)respectively; in fact

F−1(∞) = (S R)−1(∞) = R−1(S−1(∞)) =

⋃

α∈S−1(∞)

R−1(α)

So if S−1(∞) = α we can choose a rational function λ ∈ k(t) of degree 1 such thatλ(α) =∞ (for example if α ∈ k we can choose λ(t) = 1/(t− α); if α =∞ then s and rare polynomials). Hence the rational function s′ = s λ−1 satisfies s′−1(∞) = ∞ andso it is a polynomial. For the same reason r′ = λ r is a polynomial and we have thatf = s′ r′. Obviously k(r) = k(r′) since λ has degree 1.

1.2 Minimal couples of rational functions

Let k be a fixed field and consider the pure trascendental field k(t). If (ϕ(t), ψ(t)) isa couple of rational functions of k(t) then, since the trascendence degree of k(t) over k isone, they are algebraically dependent in k(t), which means that there exists a polynomialF (X,Y ) ∈ k[X,Y ] such that the following equality holds in k(t)


F (ϕ(t), ψ(t)) = 0 (1.1)

By proposition 3.1.2 there is a unique minimal irreducible polynomial F (X,Y ) mo-dulo constant factor such that (1.1) holds. We call this polynomial minimal polyno-mial of (ϕ(t), ψ(t)).

Lemma 1.2.1 Let ϕ,ψ ∈ k(t) be such that k(ϕ,ψ) = k(t) and f ∈ k[X,Y ] be irreduciblesuch that f(ϕ(t), ψ(t)) = 0. Then deg(ϕ) = degY (f) and deg(ψ) = degX(f).

Proof : Given a rational function ϕ ∈ k(t) the degree of the field extension k(ϕ) ⊂k(t) is equal to the degree deg(ϕ) by lemma 1.1.2. The field k(ϕ,ψ) is an algebraicfunction field in one variable since x = ϕ(t), y = ψ(t) are algebraically dependent; sincef is irreducible, the polynomial f(x, Y ) ∈ k(x)[Y ] is the minimal polynomial of y overthe field k(x) and so the degree of k(x, y) over k(x) is equal to the degree in Y of f .By hypothesis we have k(x, y) = k(t) from which the thesis follows immediately for therational function ϕ. Symmetrically we conclude for ψ.

¿From now on we call minimal a couple of rational functions (ϕ,ψ) which satisfiesthe hypothesis of the lemma; from a geometric point of view this means that the mapF : A1 → C, given by t 7→ (ϕ(t), ψ(t)), where C is the image curve of the application F(defined as the zero-locus of the minimal polynomial f of (ϕ,ψ)), has degree one, thatis a birational map (see chapter 4).

In general if a couple (ϕ,ψ) of rational functions is not minimal then its miminalpolynomial f(X,Y ) satisfies an algebraic equation f(ϕ1(t), ψ1(t)) = 0 where (ϕ1, ψ1) isminimal: this follows by Luroth theorem (see 1.1.3).

In fact by this theorem, the field generated by ϕ and ψ over k, that is k(ϕ(t), ψ(t))which is a subfield of k(t), is equal to k(η(t)) for a certain rational function η ∈ k(t); thecouple (ϕ,ψ) is minimal exactly in the case when the degree of η is equal to one: in thiscase it follows that k(η) = k(t) by lemma 1.1.2.

If the degree of h = deg(η) is greater than one we have that ϕ = ϕ1 η andψ = ψ1 η for certain rational functions ϕ1 and ψ1 of degree less than the degreesof ϕ and ψ respectively. If an irreducible polynomial f ∈ k[X,Y ] satisfies the relationf(ϕ1(η), ψ1(η)) = 0 (η is seen as a trascendental element over k) then we also have thatf(ϕ1(η(t)), ψ1(η(t))) = 0; conversely if we have f irreducible such that

f(ϕ1(η(t)), ψ1(η(t))) = 0

then for the surjectivity of rational functions of P1(k) = k ∪ ∞ in itself (given byrational functions in k(t)) it follows that f(ϕ1(η), ψ1(η)) = 0.

Observe that k(ϕ1, ψ1) = k(η) (ϕ1 and ψ1 are rational functions of k(η)). Hence wehave proved the following lemma, which generalizes the previous one:

Lemma 1.2.2 Let ϕ,ψ, η ∈ Q(t) be such that k(ϕ,ψ) = k(η) ⊂ k(t). Let f ∈ k[X,Y ] beirreducible such that f(ϕ(t), ψ(t)) = 0; then deg(ϕ) = h ·degY (f) e deg(ψ) = h ·degX(f)where h = [k(t) : k(η)].

1.3. VALUATION RINGS 13

k(t)

h

n

====

====

====

====

==

m

k(ϕ,ψ) = k(η)

degY (f) NNNNNNNNNNN

degX(f)ppppppppppp

k(ϕ) k(ψ)

The proof of this lemma also shows that we can associate to each couple of ratio-nal functions (ϕ,ψ) a minimal couple (ϕ1, ψ1) of rational functions which defines thesame curve C = (x, y) ∈ A2(k)|f(x, y) = 0. We also say that (ϕ1, ψ1) is a minimalparametrization of the curve C.

The following lemma is an immediate consequence of the previous results.

Lemma 1.2.3 Let ϕ,ψ ∈ k(t). If (degϕ, degψ) = 1 then (ϕ,ψ) is minimal.

1.3 Valuation rings

Valuation rings of algebraic function fields in one variable are the algebraic counter-part of geometric points of algebraic curves. More precisely the set of valuation ringsof an algebraic function fields K corresponds bijectively to the geometric point a non-singular model of a curve with rational function field isomorphic to K (see paragraph3.4 ) .

Definition 1.3.1 Let K be a field. A valuation ring of K is a subring O ⊂ K such thatfor each x ∈ K we have x ∈ O or x−1 ∈ O.

If k ⊂ K is a field extension and O is a valuation ring of K such that k ⊂ O thenwe say that O is a valuation ring of K over k.

Lemma 1.3.2 Valuation rings are local rings integrally closed.

Proof : The ideal P = x ∈ O|x−1 6∈ O is maximal and O − P = O∗.If F is the quotient field of O (O is a domain because it is contained in a field) and

x ∈ F is integral over O, then

xn + an−1xn−1 + . . .+ a0 = 0

with ai ∈ O. If x 6∈ O then x−1 ∈ O. If we multiply the previous equality by x−n weobtain

1 = −an−1x−1 − . . .− a0x

−n


Hence we have a contradiction: the right member is in O so 1 would be an elementof O. .

Let K/k be an algebraic function field in one variable. If O is a valuation ring ofK over k with maximal ideal P then k ⊂ O/P since k ∩ P = 0; the field O/P iscalled residue field of the valuation ring O. The following proposition shows that theextension k ⊂ O/P is finite (for a proof of this fact see Rosen, [36] chap. 5, pg. 46):

Proposition 1.3.3 Let K/k be an algebraic function field in one variable and let (O,P )be a valuation ring of K over k; then the residue field of O is a finite extension of k.

If we assume that the base field k is algebraically closed then the residue field ofvaluation rings are all equal to k.

The next result classifies valuation rings of a purely trascendental extension oftrascendence degree one, which is, as we said above, the ”simplest” case of algebraicfunction field (geometrically it corresponds to the curve P1). These field has two typeof valuation rings: the localization of k[t] with respect to a prime ideal of k[t] and thevaluation ring at infinity, which is the ring of rational functions which are regular atinfinity, that is the rational functions f(t)/g(t), with deg(g) ≥ deg(f).

Proposition 1.3.4 Let K = k(t) be a purely trascendental extension of degree 1 and letO be a valuation ring of K over k; then either O is the localization of k[t] for a certainprime ideal p of k[t] or O = k[1/t](1/t).

Proof : Let P = x ∈ O|x−1 6∈ O be the maximal ideal of O.If t ∈ O then k[t] ⊂ O since O is a ring. The ideal p = P ∩ k[t] is a prime ideal

of k[t], different from the zero ideal (otherwise O = K); let f ∈ k[t] be an irreduciblepolynomial which generates the ideal p. If x ∈ k[t]p then x = h(t)/g(t) with h, g ∈ k[t]coprime and g 6∈ p ⊂M ; hence g−1 ∈ O∗ and so h/g ∈ O.

Let x be in O of the form x = h(t)/g(t), where h, g ∈ k[t] are coprime polynomialsand suppose that x 6∈ k[t]p: then x−1 ∈ pk[t]p, the maximal ideal of the localizationk[t]p; hence g ∈ p ⊂ P and h 6∈ p. So h ∈ O∗ and 1/h · x = 1/g ∈ O. Since g ∈ P thisleads to a contradiction. So if t ∈ O we have proved that O = k[t]p, where p is a primeideal of k[t].

If t 6∈ O then s = t−1 ∈ P ; by considering again the prime ideal p = P ∩ k[s] we havethat s ∈ p and so s is a generator of p (its degree is the least possible). We concludethat O = k[s]s = k[1/t]1/t. .

In particular we observe that if k is algebraically closed then the valuation rings ofk(t) are in bijection with the elements of k ∪ ∞. From a geometric point of view thevaluation ring Op of k(t), for p ∈ k, is the set of rational functions ϕ(t) = f(t)/g(t) suchthat g(p) 6= 0, that is ϕ does not have a pole in p; the valuation ring at infinity O∞ is theset of rational functions ϕ(t) = f(t)/g(t) such that deg(g) ≥ deg(f) (which implies thatϕ does not have a pole at infinity). Observe that maximal ideals of valuation rings of k(t)

1.3. VALUATION RINGS 15

are principal ideal; this particular kind of valuation ring is called discrete valuationring (DVR for short, see Serre [40]). Actually valuation rings of algebraic function fieldsin one variable are discrete valuation rings (see next results).

The next result can be seen as a particular case of Noether normalization lemma,valid in the case of separable extension (and hence simple extension if they are finite).

Lemma 1.3.5 Let E = K(α) be a finite extension of a field K and A a subring of Ksuch that K is the quotient ring of A. Then there exists α′ ∈ E such that E = K(α′)and α′ is integral over A.

Proof : By hypothesis α is a root of a monic polynomial:

f(X) = Xn + an−1Xn−1 + . . .+ a1X + a0

where ai ∈ K. Since K is the quotient field of A then there exists d ∈ A such thatdai ∈ A for every i = 0, . . . , n−1. If we multiply the equation f(α) = 0 by dn we obtain:

(dα)n + dan−1(dα)n−1 + . . .+ dn−1a1(dα) + dna0 = 0

If we define α′ = dα we obtain the desired element.

If A is a subring of a field K, the integral closure of A in K is defined to be the setof all elements of K which are integral over A (i.e. they satisfy a monic equation overA). Next proposition (see [9]) characterizes integral closure in terms of valuation rings.

Proposition 1.3.6 Let A be a ring and let K be a field such that A ⊂ K. Then theintegral closure of A in K is equal to the intersection of all the valuation rings of Kwhich contain A.

Proof : Let O be a valuation ring of K which contains A; it is integrally closed bylemma 1.3.2. If an element x ∈ K is integral over A then it is integral over O and so itbelongs to O.

Conversely let x ∈ K be such that it is not integral over A: then it is clear thatx 6∈ A[ 1

x ] = A[y]. Since y is not invertible in A[y] there exists a maximal idealM of A[y]which contains y. By a classic theorem of Chevalley (see [9], chap.1 §.4) there exists avaluation ring (O,P ) of K which contains the ring A[y] and such that P ⊂M; since themaximal ideal of a valuation ring is equal to the set α ∈ O|α−1 6∈ O we immediatelysee that x 6∈ O. The proof is complete.

This last lemma prove that valuation rings of algebraic function fields are discretevaluation rings, as we said before.

Lemma 1.3.7 Let k be a perfect field and K an extension of k of trascendence degreeone. Then the valuation rings of K are discrete valuation rings.


Proof : Let x ∈ K be a trascendental element over k; then there exists y ∈ Kalgebraic over k(x) such that K = k(x, y). The intersection of a valuation ring O of Kwith k(x) is a discrete valuation ring (see above; this is true also in the case of k notalgebraically closed: O ∩ k(x) = k[x]p, where p ⊂ k[x] is a prime ideal). The extensionof discrete valuation rings in finite algebraic extension are discrete valuation rings (seeLang, [26]); so O is a discrete valuation ring.

Chapter 2

Ritt’s decomposition theorem for

polynomials

In this chapter we want to expose an article of Ritt of 1922, ”Prime and composite

polynomials” (see [35]), which deals with decomposition of polynomial with respect tothe operation of composition of functions. Strictly speaking Ritt proved that there is asort of factorization in terms of indecomposable polynomials (i.e. polynomials f ∈ C[X]of degree greater than one such that there do not exist non-linear g, h ∈ C[X] such thatf = g(h(X))): the number of terms of a maximal decomposition is unique and the orderof the terms of two maximal decomposition are the same up to the order.

¿From this article many others arise (see [11], [12], [29], among the others); theydo not add anything new but they prove the same result with other methods, such asramification theory of valuations in algebraic function fields.

2.1 Introduction

We work with the monoid (C(X), ) and the submonoid (C[X], ), the field of ra-tional functions and the ring of polynomials over C, where denotes the operation ofcomposition of functions; this operation will be denoted in the following ways

f g(X) = f(g(X))

where f and g are two rational functions.

Lemma 2.1.1 If f(X) and g(X) are rational functions, then deg(fg) = deg(f) deg(g).

The following lemma describes the units of (C(X), ).

Lemma 2.1.2 (C(X), )∗ = f(X) ∈ C(X)|deg(f) = 1 = ax+bcx+d ∈ C(X)|ad − bc 6=0 ∼= PSL(2,C).

17

18 CHAPTER 2. RITT’S DECOMPOSITION THEOREM FOR POLYNOMIALS

Definition 2.1.3 A rational functions f(X) is called indecomposable (prime in theoriginal article of Ritt) if its degree is strictly larger than one and there do not existrational functions g(X), h(X) of degree strictly less of deg(f) such that

f(X) = g h(X)

It is natural to study how a rational function decomposes into indecomposable func-tions; by induction any rational functions of degree more then one can be written as thecomposition of indecomposable functions. A decomposition of a rational function

F (X) = f1 . . . fn(X)

is called maximal if the rational functions fi are indecomposable; the rational func-tions fi of this decomposition are called components of the decomposition of F .

Proposition 2.1.4 Let F ∈ C(X). Then there is a bijection between the set of decom-position of F and chain of subfield between C(F ) and C(X).

Maximal decompositions correspond to maximal chain of fields.

Proof : Let F = f1 . . .fr be a decomposition. We associate to this decompositionthe chain of field C(F ) ⊂ C(f2 . . . fr) ⊂ . . . ⊂ C(fr) ⊂ C(X).

The opposite map is defined in this way: if C(F ) ⊂ K1 ⊂ . . . ⊂ Ks ⊂ C(X) is achain of subfields then Ks = C(fs) , Ks−1 = C(fs−1(fs)), Ki = C(fi(fi+1(. . . (fs)))) fori = 1, . . . , s − 2, by Luroth’s theorem. So we have the following decomposition of F :F = f1 . . . fs.

Obviously these maps are inverse to each other.The other statement follows immediately. .

Corollary 2.1.5 Let f ∈ C(X). The numbers of maximal non-equivalent decomposi-tions is finite.

Proof : In fact the extension C(f) ⊂ C(X) is separable so the numbers of intermediatefields between C(f) and C(X) is finite. .

Lemma 2.1.6 Let F ∈ C[X] and let F = f g be a decomposition of F , where f and gare in C(X).

Then there exists a linear function µ ∈ C(X) such that f µ and µ−1 g are polyno-mials.

This lemma implies that we can assume that (indecomposable) factors of a polyno-mial are themselves polynomials.

Proof : The proof follows from the polynomial version of Luroth’s theorem; considerthe field extension C(F ) ⊂ C(g) ⊂ C(X): by theorem 1.1.4 there exists a polynomial

2.1. INTRODUCTION 19

g′ ∈ C[X] such that C(g) = C(g′). Then F = f ′ g′ and it is immediate to see thatf ′−1(∞) = ∞ so f ′ is a polynomial.

This lemma is straightforward.

Lemma 2.1.7 If f g = h g then f = h.

In the case of a polynomial, Ritt’s work shows that the decomposition is not necessa-rily unique but the degree of the indecomposable factors are unique up to permutationand also the number of indecomposable components of a maximal decomposition doesnot depend on the maximal decomposition.

Ritt’s work in [35] for decomposition of polynomials contains the following two the-orem, now known as first and second theorem of Ritt.

Theorem 2.1.8 (First theorem of Ritt) Let f(X) be a polynomial with complexcoefficients and let

f(X) = f1 . . . fr(X)

f(X) = g1 . . . gs(X)

be two maximal decomposition of f , where fi(X)i=1...r and gj(X)i=1...s are indecom-posable polynomials.

Then r = s and deg(fi)i=1,...,r = deg(gj)j=1,...,s.

So the number of indecomposable factors of a maximal decomposition of a polynomialis unique and the degrees of factors are the same up to permutation.

Two decompositions with the same numbers of components

f = f1 . . . fr = g1 . . . gr

are equivalent if there exist r − 1 linear polynomials λ1, . . . , λr−1 such that

g1 = f1 λ1, . . . gi = λ−1i−1 fi λi, . . . gr = λ−1

r−1 fr

henceg1 . . . gr = (f1 λ

−11 ) . . . (λ−1

i−1 fi λi) . . . (λ−1r−1 fr)

The first theorem of Ritt implies that this is an equivalence relation between maximaldecompositions of a polynomial F .

After that, Ritt studied the equation

ϕ α = ψ β

where ϕ, α, ψ, β are indecomposable polynomials.For example there are two cases where this equality is satisfied:

• Xn Xrg(Xn) = Xrg(X)n Xn , where g(X) is a polynomial


• Tn Tm = Tm Tn , where Tn is the n-th Tchebychev polynomial

In the case of polynomial these are the only case where maximal bidecompositionsoccur.

We will prove the following lemma.

Lemma 2.1.9 Let f be a polynomial with complex coefficients and let

f = ϕ α = ψ β

be two decompositions of f with α and β indecomposable. Then deg(α) = deg(β) or(deg(α),deg(β)) = 1.

Next theorem is now known as the second theorem of Ritt (see [38]) and it concernsmaximal bidecompositions of polynomials.

Theorem 2.1.10 (Second theorem of Ritt) If ϕ, α, ψ, β are indecomposable polyno-mials such that

ϕ α = ψ β

with deg(ϕ) = deg(β) = m, deg(α) = deg(ψ) = n and (n,m) = 1 then the decompo-sition is equivalent to one of the examples shown above.

If f ∈ C[X] let K the splitting field of C(X) over C(f); the monodromy group of fis the Galois group Gal(K/C(f)) and it is denoted with Mon(f).

Ritt’s results can be grouped together in the following theorem:

Theorem 2.1.11 Let f(X) be a polynomial with complex coefficients and let

f(X) = f1 . . . fr(X)

f(X) = g1 . . . gs(X)

be two maximal decomposition, where fi(X)i=1...r and gj(X)i=1...s are polynomials.Then r = s, deg(fi)i=1,...,r = deg(gj)j=1,...,s and Mon(fi)i=1,...,r = Mon(gj)j=1,...,s.

Moreover it is possible to pass from one decomposition to another by means of thefollowing three ways:

• fi fi+1 = (fi L) (L−1 fi+1), where L is a linear polynomial

• Xn Xrg(Xn) = Xrg(X)n Xn , where g(X) is a polynomial

• Tn Tm = Tm Tn, where Tn is the n-th Tchebychev polynomial

The result of the monodromy groups is due to Mueller (see his article [31]).Ritt’s work about decomposition of a polynomial function was the study of mono-

dromy groups of a polynomial f ∈ C[X] and the link between the monodromy groupsof the components of a maximal decomposition of f . We will show a proof of the firsttheorem of Ritt; in the next paragraphs we will recall some facts about coverings oftopological spaces and theory of blocks which arise in the action of a finite group on afinite set. We will need these two theories in the proof of Ritt’s results.

2.2. MONODROMY AND GALOIS COVERING 21

2.2 Monodromy and Galois covering

In this section we recall some basic facts about coverings (for more references see [19]and [30]).

Let X and Y be topological spaces; a covering from X to Y is a continue andsurjective map ϕ such that for each y ∈ Y there exists an open neighbourhood V ⊂ Yof y such that the preimage of V is a disjoint union of open sets Ui of X which arehomeomorphic via ϕ to V , that is ϕ|Ui

: Ui → V is a homeomorphism. The covering isfinite if for each y in Y the fiber ϕ−1(y) is finite; if we assume, as we will do from nowon, that Y is connected then the cardinality of the fibers of a finite covering is constantand it is called the degree of the covering. If we want to specify the base points of thecovering, that is elements x0 ∈ X and y0 ∈ Y such that ϕ(x0) = y0, we will use theclassical notation ϕ : (X,x0)→ (Y, y0).

If ϕ : (X,x0)→ (Y, y0) is a covering, the following map between fundamental groupsof X and Y is well defined

ϕ∗ : π1(X,x0) → π1(Y, y0)[γ] 7→ [ϕ(γ)]

where [γ] is the homotopy class of a closed path γ in X with base point x0 andin the same way [ϕ(γ)] (we will omit the square brackets); by monodromy lemma thehomomorphism of groups ϕ∗ is injective (see [19]). We call characteristic subgroupthe subgroup H = ϕ∗(π1(X,x0)) of G = π1(Y, y0); the index of H in the fundamentalgroup π1(Y, y0) is equal to the degree of the covering: in fact there is a bijection betweenthe fiber ϕ−1(y0) and the set G/H (see [19]).

Two coverings ϕ1 : (U1, u1)→ (V, v0) and ϕ2 : (U2, u2)→ (V, v0) are isomorphic ifthere exists a homeomorphism φ : U1 → U2 such that ϕ2 φ = ϕ1 (note that we do notimpose that φ(u1) = u2). If U1 = U2 = U and φ : (U, u0) → (V, v0) is a covering thenwe define the group Deck(φ), the group of automorphisms of the covering

Deck(φ) + φ : U → U |ϕ φ = ϕ

It is easy to check that every element of Deck(φ) preserves the fibers of ϕ, that isφ(ϕ−1(v)) = ϕ−1(v), for all v ∈ V .

Proposition 2.2.1 Two coverings of a topological space (V, v0) are isomorphic if andonly if their characteristic subgroups are conjugate in π1(V, v0).

For a proof of this proposition see [19].

Theorem 2.2.2 If (V, v0) is a topological space then there is a bijection

ϕ : (U, u0)→ (V, v0)covering/∼ → conjugacy class of H < π1(V, v0)


Proof : If H < G = π1(V, v0) we consider the universal covering U di V (see [19]);then G acts freely on U through lift of paths and V ∼= U/G. We consider the inducedaction of H on U and we obtain a covering ϕ : U/H → V which has H as characteristicsubgroup.

Conversely we associate to a covering ϕ the conjugacy class of its characteristicsubgroup.

Let ϕ : (U, u0)→ (V, v0) be a finite covering of degree n. We define the monodromyof the covering as the homomorphism induced by the action of the fundamental groupG = π1(V, v0) on the fiber ϕ−1(v0):

Φ : π1(V, v0) → Sϕ−1(v0)∼= Sn

γ 7→ ui 7→ γui(1)

where Sϕ−1(v0) is the set of permutations of the finite set ϕ−1(v0) and γuiis the

unique lifting in U of γ with base point ui, where ui belongs to the fiber ϕ−1(v0); forthe uniqueness of lifting of paths with fixed base point the application Φ is well defined(see [19]). The group Φ(π1(V, v0)) ⊂ Sn, permutation group of the set ϕ−1(v0), is calledthe monodromy group of the covering ϕ and it is denoted with Mon(ϕ); since U isconnected it follows that this group acts transitively on the fiber ϕ−1(v0) and so it is atransitive subgroup of Sn. It also acts faithfully on the set ϕ−1(v0) and it is isomorphicto π1(V, v0)/ ker(Φ).

Theorem 2.2.3 If (V, v0) is a topological space then there is a bijection

ϕ : (U, u0)→ (V, v0)covering/∼ → transitive action of π1(V, v0) on a finite set/∼

Proof : We have already seen that if ϕ : (U, u0)→ (V, v0) is a covering then π1(V, v0)acts transitively on the set ϕ−1(v0).

Conversely suppose that the group G = π1(V, v0) acts transitively on a finite set I;let H be the stabilizer of an element i ∈ I. Then by the same argument of the proof oftheorem 2.2.2 there exists a covering ϕ : (U, u0) → (V, v0) with characteristic subgroupH. The fiber ϕ−1(v0) is identified with the set I, since they are both in bijection withthe quotient G/H.

The following lemma describes the characteristic group of a covering φ : (U, u0) →(V, v0) in terms of the action of the fundamental group of V in v0 on the fiber φ−1(v0).

Lemma 2.2.4 Let ϕ : (U, u0) → (V, v0) be a covering. Then the stabilizer of u0 underthe action of the fundamental group π1(V, v0) is equal to the characteristic group of thecovering.

Next lemma describes the kernel of the monodromy map associated to a covering.


Lemma 2.2.5 If ϕ : (U, u0) → (V, v0) is a covering, Φ the monodromy and H thecharacteristic subgroup, then

ker(Φ) =⋂

γ∈π1(V,v0)

γHγ−1

Proof : The statement follows easily from previous lemma. We consider a genericpoint u on the fiber ϕ−1(v0): its stabilizer is equal to γHγ−1, where γ is a path in Vobtained as image via ϕ of a path σ in U with initial point in u0 and final point in u.The kernel of Φ is the intersection of all stabilizers of the points of the fiber ϕ−1(v0),which form a conjugacy class of subgroups in π1(V, v0).

We now give a definition: if H is a subgroup of a group G we set

coreG(H) +⋂

g∈G

gHg−1

which is the maximal normal subgroup of G contained in H. The subgroup H isnormal in G if and only if coreG(H) = H. So in the previous lemma we have thatker(Φ) = coreG(H), where G = π1(V, v0).

Let f, ϕ, ψ be coverings such that f = ψ ϕ, and consider the fiber f−1(v0) underthe action of the monodromy group of f . The following proposition permits us to provethat the map ϕ determines a decomposition in blocks of f−1(v0) (for the definition ofblocks see section 2.3).

Proposition 2.2.6 Let the following one be a diagram of finite coverings

(W,w0)ϕ

//

f %%KKKKKKKKK(U, u0)

ψ

(V, v0)

and let the following ones be the monodromy homomorphisms of f and ψ respectivelydefined before

F : π1(V, v0)→ Sf−1(v0)

Ψ : π1(V, v0)→ Sψ−1(v0)

If γ ∈ π1(V, v0) thenϕ F (γ) = Ψ(γ) ϕ

that is the following diagram is commutative

f−1(v0)F (γ)−−−−→ f−1(v0)

ϕ

yyϕ

ψ−1(v0)Ψ(γ)−−−−→ ψ−1(v0)


Proof : If ψ−1(v0) = u0, . . . , um−1 then f−1(v0) = B0 ∪ . . . ∪ Bm−1 where Bi =

z ∈ f−1(v0)|ϕ(z) = ui = z(i)j |j = 1, . . . , n.

Let γ ∈ π1(V, v0) and F (γ)(z(i)j ) = γ

z(i)j

be the unique lifting of γ via f in W with

initial point z(i)j .

Observe that γ + ϕ(γz(i)j

) is a path in U with initial point ϕ(z(i)j ) = ui such that

ψ(γ) = γ and so it is the unique lifting γuiof γ via ψ in U with initial point ui, that is

γ = Ψ(γ)(ui).Hence

ϕ(F (γ)(z(i)j )) = ϕ(γ

z(i)j

(1)) = γui(1) = Ψ(γ)(ui) = Ψ(γ)(ϕ(z

(i)j ))

⇒ ϕ F (γ) = Ψ(γ) ϕ

which proves the proposition.

A consequence of this theorem is the following corollary:

Corollary 2.2.7 Let the following one be a diagram of finite coverings

(W,w0)ϕ

//


ψ

(V, v0)

For each u ∈ ψ−1(v0) we set Bu + z ∈ f−1(v0)|ϕ(z) = u0; then the sets Bu, foru ∈ ψ−1(v0), are transitively permuted by the group G = π1(V, v0).

Moreover we have that H = GBu0, where H = ψ∗(π1(U, u0)) and GBu0

is the stabilizerin G of the set Bu0.

Proof : The sets Bu, for u ∈ ψ−1(v0), satisfy this property: for all σ ∈ G we haveeither σ(Bu) = Bu or σ(Bu) ∩ Bu = ∅ (where σ(Bu) has to be intended as F (σ)(Bu)).This fact implies that G acts on the set B = Buu∈ψ−1(v0).

In fact suppose that σ(Bi) ∩ Bi 6= ∅, that is for some z ∈ Bi the element σ(zi) is inBi; then by proposition 2.2.6

ui = ϕ(F (σ)(z)) = Ψ(σ)(ϕ(z)) = Ψ(ui)

Then if z′ is in Bi we have

ϕ(F (σ)(z′)) = Ψ(σ)(ϕ(z′)) = Ψ(ui) = ui

so σ(z′) is inBi. The set B is permuted transitively byG, since f−1(v0) =⋃u∈ψ−1(v0)Bu

and G acts transitively on f−1(v0).


For the second statement, H is the stabilizer in G of the point u0. If σ ∈ H thenσ(u0) = Ψ(σ)(u0) = u0. So for z ∈ Bu0 we have

ϕ(F (σ)(z)) = Ψ(σ)(ϕ(z)) = Ψ(σ)(u0) = u0

hence σ(Bu0) = Bu0 . Conversely let σ ∈ Bu0 ; then

Ψ(σ)(u0) = Ψ(σ)(ϕ(z)) = ϕ(F (σ)(z)) = ϕ(z′) = u0

where z, z′ are elements of Bu0 .

Proposition 2.2.8 Let the following one be a diagram of finite coverings

(W,w0)ϕ

//


ψ

(V, v0)

and let the following ones be the natural maps defined above

ψ∗ : π1(U, u0) → π1(V, v0)

F : π1(V, v0)→ Sf−1(v0)

Φ : π1(U, u0)→ Sϕ−1(u0)

Let Bu0 = w ∈ f−1(v0)|ϕ(w) = u0; then for each γ ∈ π1(U, u0) it follows that:

F ψ∗(γ)|Bu0= Φ(γ)

Proof : By previous corollary the stabilizer of Bu0 in G = π1(V, v0) is equal toH = ψ∗(γ)|γ ∈ π1(U, u0); so if γ ∈ π1(U, u0) then γ = ψ∗(γ) ∈ H. Hence F (γ) is apermutation of the set Bu0 . The statement says that the action of F (γ) on Bu0 is thesame of the action of Φ(γ) on the same set.

Let z be an element of Bu0

Φ(γ)(z) = γz(1)

F (γ)(z) = γz(1)

Since f = ψ ϕ it follows that the lifting γz of γ via ϕ with initial point z coincideswith the lifting of γz of γ via f with initial point z, so they have the same ending point.

We end this section with the following diagram of finite coverings (we will have todo with this situation later in the proof of Ritt’s theorem):


(W,w0)ϕ

//


ψ

(V, v0)

We set G = π1(V, v0), J = π1(U, u0) and H = π1(W,w0). Then we have H < J < G(viewing all the groups inside the others through monodromy lemma).

The monodromy groups of the three coverings are equal toMon(f) = G/coreG(H)Mon(ϕ) = J/coreJ(H)Mon(ψ) = G/coreG(J)

2.2.1 Ramified coverings

A ramified covering is a continuos, open and surjective map φ : X → Y betweentopological spaces such that there exists a discrete subset ∆ ⊂ Y such that the map

φ|X−φ−1(∆) − φ−1(∆)→ Y −∆

is a covering. The characteristic subgroup of a ramified covering is the characteristicsubgroup of the associated covering. The set ∆ is called the ramification points of theramified covering φ.

Examples of ramified covering are holomorphic maps between compact Riemannsurfaces.

If φ : X → Y a ramified covering we define the monodromy of φ in the followingway: if ∆ ⊂ Y is the set of ramification points of φ we consider the covering

ϕ + φ|U : U → V

where U + X − φ−1(∆) and V + Y −∆. We fix a base point v0 in V and we definethe monodromy of φ as the monodromy of ϕ. We will assume that U is connected.

Consider the case of a polynomial

F (Z, T ) = an(Z)Tn + an−1(Z)Tn−1 + . . .+ a1(Z)T + a0(Z) ∈ C[Z, T ]

which determines the algebraic function T over C(Z) and let π : C → P1(C) bethe associated ramified covering of compact Riemann surfaces, where C is the compactRiemann surface associated to the polynomial F (Z, T ) (roughly is the desingularizationof the curve (z, t) ∈ A2|F (z, t) = 0 ).

The monodromy is both the action of the fundamental group G = π1(P1(C)−∆, z0)

on the fiber of z0 of the covering π : C − π−1(∆) → P1(C) − ∆ where ∆ is the finite


set of ramification points of π, and also the action of G on the set of the n branches ofthe algebraic function T in a neighbourhood of z0. The group G is known to be finitelygenerated; we fix a system of generators in this way: let k = #∆ and for each zi ∈ ∆let σi be a closed path based on z0 which turns around zi and no other zj ∈ ∆ − zi,without intersecting the other paths σj .

The following propositions describes the fundamental group of a the Riemann spherewith a finite number of holes (see [44]):

Proposition 2.2.9 If ∆ is a finite subset of P1(C) and z0 ∈ P1(C)−∆ then the funda-mental group π1(P

1(C) − ∆, z0) is generated by the paths σ1, . . . , σk defined above withthe only relation

∏i=1,...,k σi = 1.

Proposition 2.2.10 Let F (Z, T ) ∈ C[Z, T ] be a polynomial of degree n in T and π :C → P1(C) be the associated covering of compact Riemann surfaces; if z0 ∈ P1(C)−∆ isa fixed point, we consider Φ : π1(P

1(C)−∆, z0)→ Sn the monodromy of the covering π.Let z ∈ ∆ and π−1(z) = p1, . . . , ps, where s < n and let mi be the ramification indexmultpi

(π) of π in pi, for i = 1, . . . , s.If σz is a generator of the fundamental group π1(P

1(C)−∆, z0) which turns aroundz then

Φ(σz) = (t1, . . . , tm1) . . . (tn−ms , . . . , tn)

that is the permutazion associated to σz of the branches of algebraic function of Fdefined locally in z0 decompones in s cycles, each of them of lenght mi.

Lemma 2.2.11 If F : X → Z, Φ : X → Y and Ψ : Y → Z are ramified coveringsbetween compact Riemann surfaces such that F = Ψ Φ then ∆F = ∆Ψ ∪Ψ(∆Φ)

Proof : Since

F−1(z0) = Φ−1(Ψ−1(z0)) =⋃

y∈Ψ−1(z0)

Φ−1(y)

it follows that

#F−1(z0) =∑

y∈Ψ−1(z0)

#Φ−1(y)

¿From this equality we can easily deduce the statement of the lemma. .

If f ∈ C(Z) we denote with Φf the monodromy associated to the ramified covering

f : P1(C)x → P1(C)tx 7→ t = f(x)

and with Mon(f) the monodromy group of the covering. We will prove that if we passto the extension of algebraic function fieldsM(P1(C)t) ⊂M(P1(C)x) that is C(t) ⊂ C(x)


which has degree equal to deg(f) as we have already seen, its Galois closure has Galoisgroup isomorphic to Mon(f).

2.2.2 Galois coverings

If ϕ : (U, u0)→ (V, v0) is a covering, besides the transitive action of the fundamentalgroup π1(V, v0) on the fiber ϕ−1(v0), we also have the action of the group Deck(ϕ) ofautomorphisms of the covering on the same set, since every automorphism preserves thefibers; such two group actions commute, that is

g(γ(u)) = γ(g(u))

for each g ∈ Deck(ϕ), γ ∈ π1(V, v0) and u ∈ ϕ−1(v0) (see [44]).The action of Deck(ϕ) is free, that is the stabilizers of the points of the fiber are

trivial (this follows from the uniqueness of liftings).The following proposition shows the relation between the group of automorphisms

of a covering and the characteristic group of the covering (see [19] and [30]).

Proposition 2.2.12 If ϕ : (U, u0) → (V, v0) is a covering and Deck(ϕ) is the group ofautomorphisms, then Deck(ϕ) is isomorphic to NG(H)/H, where G = π1(V, v0), H isthe characteristic subgroup and NG(H) is the normalizer of H in G.

A covering ϕ : (U, u0) → (V, v0) is called a Galois covering if the group Deck(ϕ)acts transitively on the fibers of the covering; in this case it is immediate to prove thatthe cardinality of Deck(ϕ) is equal to the cardinality of the fiber ϕ−1(v0) since the actionis free. The following lemma characterizes Galois coverings in terms of the characteristicsubgroup.

Lemma 2.2.13 Let ϕ : (U, u0) → (V, v0) be a covering; then ϕ is a Galois covering ifand only if the characteristic subgroup H ∼= π1(U, u0) is normal in G = π1(V, v0). In thiscase the group of the automorphisms Deck(ϕ) is isomorphic both to the quotient groupG/H and to the monodromy group of the covering.

Proof : If H⊳G then by previous propositon Deck(ϕ) acts transitively on the fibers,since G acts transitively on fibers.

Conversely, if ϕ is a Galois covering, since Deck(ϕ) acts freely we have that its order isequal to the degree n of the covering. But n = #G/#H and #Deck(ϕ) = #NG(H)/#H;it follows that #G = #NG(H), so H is normal in G.

The last statement follows both from previous proposition and from lemma 2.2.5.

So characteristic subgroups of Galois coverings are equal to the kernel of the mono-dromy homomorphism; we remind that in general the equality of lemma 2.2.5 holds.

We note that if ϕ : (U, u0)→ (V, v0) is a Galois covering then V ∼= U/Deck(ϕ), whereDeck(ϕ) is the group of automorphisms of the covering (see [19]).


Definition 2.2.14 Let ϕ : (U, u0) → (V, v0) be a finite covering. The Galois closureof ϕ is defined as the Galois covering ψ : (W,w0)→ (U, u0) with characteristic subgroupK = ker(Φ) ⊳ π1(U, u0), where Φ is the monodromy homomorphism of ϕ.

Note that φ = ψ ϕ : (W,w0) → (V, v0) is a Galois covering since its characteristicsubgroup K is normal in π1(V, v0). Observe also that Mon(ϕ) is equal to Mon(φ) andthat this group acts transitively and faithfully on the set ϕ−1(v0) and it acts transitivelyand freely on the set φ−1(v0).

We have the following commutative diagram

(W,w0)ψ

//

φ %%KKKKKKKKK(U, u0)

ϕ

(V, v0)

We can also define the Galois closure of ϕ as the Galois covering φ : (W ′, w′0) →

(V, v0) with characteristic subgroup K = ker(Φ). Let H be the characteristic subgroupof ϕ (isomorphic to the fundamental group of (U, u0)) and let G be the fundamentalgroup of (V, v0); since K < H < G then (W ′, w′

0)∼= (W,w0) because they are coverings

of (V, v0) (via φ and ϕ ψ respectively) with the same characteristic subgroup (seeproposition 2.2.1).

Moreover the monodromy group of ϕ, Mon(ϕ), is isomorphic to the group of auto-morphisms of the covering φ = ϕ ψ.

In the case of a finite ramified covering f : X → Y of compact Riemann surfacesin order to define the Galois closure we consider first the finite covering f ′ : U → Vobtained by removing the ramification values and their fibres as already seen; then weconsider the Galois closure of f ′, that is g : W → U . By the following proposition (see[28]) there exists a compact Riemann surface Z such that W → Z is an embedding withZ −W finite and there exists a ramified covering G : Z → X that extends the coveringg (G|W = g), that is the following diagram is commutative (we omit the base points)

W //

g

φ

Z

G

U //

f ′

X

f

V // Y

Proposition 2.2.15 Let Y be a compact Riemann surface and P ⊂ Y be a finite set.If f : U → Y − P is a finite covering then there exists a compact Riemann surface X, abiholomorphic inclusion i : U → X and a holomorphic map F : X → Y (that is a finiteramified covering) such that F i = j f , where j : Y −P → Y is the natural inclusion.


Ui //

f

X

F

Y − Pj

// Y

Lemma 2.2.16 Let ϕ : U → V a covering with characteristic subgroup H and ψ : W →U its Galois closure. If φ : W → V is a Galois covering such that its characteristicsubgroup is contained in H, then there exists a covering θ : W → W such that φ =ϕ ψ θ.

W

φ

111111111111111

θ //W

ψ

U

ϕ

V

Proof : The statement follows from the fact that K = ker(Φ) is the maximumsubgroup of H which is normal in π1(V ).

Theorem 2.2.17 Let f(T ) ∈ C(T ) be a rational function.Let K be the Galois closure of the finite algebraic extension of fields C(Z) ⊂ C(T ),

where Z = f(T ) and let G be the Galois group of the extension K/C(Z).Let G be the group of automorphisms of the Galois closure of the finite covering

determined by the rational function f

f : P1(C)→ P1(C)

t 7→ x = f(t)

.Then G is isomorphic to G. In particular we have that LG = C(f).

Proof : The group G is isomorphic to the monodromy group of f , as we have alreadyobserved.

We have the following extension of fields

C(Z) ⊂ C(T ) ⊂ K

We also have the following coverings of compact Riemann surfaces

P1(C)z ← P1(C)t ← C


where C is the compact Riemann surface (unique up to biolomorphisms) which is theGalois closure of P1(C)t → P1(C)z.

Obviously we have thatM(P1(C)z) = C(Z) andM(P1(C)t) = C(T ).

We want to show that the field K ′ = M(C) of the meromorphic functions of C isisomorphic to the field K. In fact let C be the compact Riemann surface associated tothe algebraic function field K; we have the following maps

C → P1(C)t → P1(C)z

such that C → P1(C)z is a Galois covering (see [44], theorems 5.9 and 5.12) and thekernel of its monodromy homomorphism is contained in the characteristic subgroup ofthe covering P1(C)t → P1(C)z; by proposition 2.2.16 we have the following maps

C → C → P1(C)t → P1(C)z

and the following inclusion of fields

C(Z) ⊂ C(T ) ⊂ K ′ ⊂ K

since K ′/C(Z) is a Galois extension (see [44]) it follows that K = K ′ (K is thesplitting field of f(T )−Z over C(Z) and K ′ contains a root of this polynomial, since itcontains C(T )). Hence C = C.

We define the following map between G and G

Ψ : G → Gφ 7→ Ψ(φ) + λ 7→ λ φ−1

By theorem 5.9 of [44] (which uses a non-trivial result like Riemann’s existencetheorem) it is an isomorphism of groups.

Corollary 2.2.18 Let f ∈ C(T ) be a rational function such that f = ψϕ, where ψ,ϕ ∈C(T ). We have the following diagram of finite coverings (once removed ramificationpoints):

C

g

(W,w0)

f

ϕ

yyttttttttt

(U, u0)

ψ %%KKKKKKKKK

(V, v0)


where g : C → (W,w0) is the Galois closure of f : (W,w0) → (V, v0). Set G =π1(V, v0), J = π1(U, u0), H = π1(W,w0) and K = π1(C, c0).

Then Deck(g) ∼= H/K and Deck(ϕ g) ∼= J/K.

Proof : The proof follows immediately from the fact that g : (C, c0)→ (W,w0) andϕ g : (C, c0)→ (U, u0) are Galois covering, since K ⊳H and K ⊳ J (remember that Kis normal in G). We conclude by applying lemma 2.2.13. .

In the proof of Ritt’s theorem we will pass from the algebraic point of view (theextension of algebraic function fields C(Z) ⊂ C(T ) ⊂ L) to the topological point of view(the finite ramified coverings C → P1(C)t → P1(C)z) without problems.

2.3 Blocks

Let G be a group acting on a finite set Ω: a block or G-block is a subset B of Ωsuch that for all g ∈ G we have either g(B) = B or g(B) ∩ B = ∅. The sets Ω, ∅, α,where α ∈ Ω, are examples of blocks; these are called trivial blocks.

The group G is called primitive if there are no non-trivial blocks; otherwise it iscalled imprimitive. If G is imprimitive then the set ΩB = g(B)|g ∈ G is calledfundamental system of blocks, where G acts transitively as a permutation group; wedenote GB the subgroup of G which stabilizes B, that is the set of g ∈ G such thatg(B) = B. If B is a block then the set GBi

|Bi ∈ ΩB is a conjugacy class of subgroupsof G.

The intersection of two G-blocks is a G-block. If G acts transitively on Ω then theunion of blocks of a fundamental system of blocks is equal to Ω, hence in this case #Bdivides #Ω.

We remind that the action of a group is free if all the stabilizers are trivial, and it isfaithful if g(α) = α for all α ∈ Ω implies that g = Id.

If B is a G-block and α ∈ B then Gα ⊂ GB, where Gα is the stabilizer of α.

The following lemma describes the structure of blocks in the particular case of acyclic group of order n acting on a finite set of order n.

Lemma 2.3.1 Let G be a cyclic group of order n which acts transitively over a finiteset Ω of n elements. Then for each divisor d of n there exists exactly one fundamentalsystem of blocks ΩB of Ω, whose blocks have cardinality d and #GB = d. Moreover ifd1|d2|n then Bd1 ⊂ Bd2 and

Bd2 =⋃

g∈GBd2

g(Bd1) =⋃

g∈GBd2/GBd1

g(Bd1)

2.3. BLOCKS 33

Proof : Observe that the action is free since G and Ω have the same cardinality.

The set Ω = x0, . . . , xn−1 is in bijection with G =< g >= 1, g, . . . , gn−1 via themap Ψ : G → Ω, gi 7→ gi(x0) = xi. So if n = dk and Hd =< gk > is the subgroupof G of cardinality d then the set Ψ(Hd) = x0, xk, . . . , x(d−1)k = Bd is a block of delements whose stabilizer GBd

is Hd; the conjugates of Bd under the action of G form afundamental system of blocks whose elements have cardinality d.

The second statement follows both from the structure of subgroups of a cyclic groupand from the fact that GBd1

⊂ GBd2if d1|d2|n.

Proposition 2.3.2 Let G be a group acting transitively on a finite set Ω

Φ : G → SΩ

g 7→ x 7→ g(x)

where SΩ is the permutation group of Ω. Then ker(Φ) = coreG(H), where H =StG(x0) is the stabilizer of an element x0 ∈ Ω in G ; moreover G = G/ ker(Φ) is a groupwhich acts faithfully and transitively on Ω.

If B is a fundamental system of G-blocks then G acts transitively on B:

ΦB : G → SBg 7→ B 7→ g(B)

and ker(ΦB) = coreG(K), where K = StG(B) is the stabilizer of a block B ∈ B inG. The group GB = G/ ker(ΦB) acts transitively and faithfully on B; without loss ofgenerality we may suppose that x0 ∈ B and so H ⊂ K.

The kernel of the natural projection

π : G → GB

is equal to coreG(K), where K is the image of K in G, thus StG(B), the stabilizer of Bin G.

Proof : Since G acts transitively then the set of stabilizers S = StG(x)|x ∈ Ω isa conjugacy class of subgroups of G, which means it is equal to gHg−1|g ∈ G, whereH = StG(x0) for some x0 ∈ Ω. The kernel of Φ is the subgroup

g ∈ G| g(x) = x ∀ x ∈ Ω =⋂

x∈Ω

StG(x) = coreG(H)

Observe that B is also a fundamental system of blocks for the group G.

In the same way we have ker(ΦB) = coreG(K). It is clear that the action of thegroups G and GB over Ω and ΩB respectively is faithful and transitive.

For the last statement we proceed as follows: since ker(Φ) ⊂ ker(ΦB) we have thefollowing diagram of group homomorphisms


G

π

G

π1

>>

π2 A

AAAA

AA

GB

where π1 and π2 are the canonical quotient maps, with ker(π1) = coreG(H) and ker(π2) =coreG(K). The map π is canonically defined as π(g) = π2(g) for g = π1(g) ∈ G, suchthat π π1 = π2.

We have

ker(π) = g ∈ G |π(g) = 0

= π1(g) ∈ G |π2(g) = 0

= π1(g) ∈ G | g ∈ ker(π2)

= π1(coreG(K))

Since coreG(K) =⋂g∈G gKg

−1 then π1(coreG(K)) =⋂g∈G gπ1(K)g−1 (if p : G →

G/H is a group homomorphism and H < A,B < G are subgroups, then p(A ∩ B) =p(A) ∩ p(B)).

Hence ker(π) = π1(coreG(K)) = coreπ1(G)(π1(K)).

A first remarkable result is the following criterion which relates decomposability ofrational functions and their monodromy groups:

Theorem 2.3.3 Let f ∈ C(T ) be a rational function of degree N > 1. Then f isindecomposable if and only if Mon(f) is primitive.

Proof : We prove that f is decomposable if and only if the group G = Mon(f) isimprimitive; we remind that Mon(f) acts over the set of the N roots of F (T,Z) overC(Z), where F (T,Z) + f(T ) − Z if f is a polynomial or F (T,Z) + f1(T ) − Zf2(T ) iff = f1/f2 is a rational function.

Note that this action is transitive because F is irreducible over C(Z) and Mon(f) isisomorphic to the Galois group of F over C(Z).

Suppose that f = ψ ϕ, where ψ,ϕ are rational functions such that deg(ψ) = m > 1and deg(ϕ) = n > 1; we have the following holomorphic maps

P1(C)tϕ

//

f $$JJJJJJJJJP1(C)u

ψ

P1(C)z

2.3. BLOCKS 35

(the letters mark the variable of the ”differents” P1(C)).Let z0 ∈ P1(C) be a regular value of f (that is #f−1(z0) = deg(f)) and consider

the sets Bui= t ∈ f−1(z0)|ϕ(t) = ui, where uii=1,...,m = ψ−1(z0); by corollary

2.2.7 the sets Bui, for ui ∈ ψ

−1(z0), form a fundamental system of blocks for the groupG = π1(P

1(C)−∆), hence by proposition 2.3.2 they form a fundamental system of blocksfor the group G/(ker(F )) = Mon(f) (where F is the usual monodromy homomorphismof f). The cardinality of each block Bui

is deg(ϕ) and there are deg(ψ) of them. So thegroup G acts imprimitively on the set f−1(z0).

Suppose now that G acts imprimitively: we want to show that f is decomposable ina non-trivial way. Let z0 be a regular value of f and t1, . . . , tN be the fiber of z0.

If B1 = t1, . . . , tn is a non-trivial G-block (1 < n < N), consider the fundamentalsystem of blocks B = Bi = g(Bi)|g ∈ G; trivially this is also a fundamental system ofblocks for the group G, since G = G/ ker(Φ) and ker(Φ) is the kernel of the action of Gon f−1(z0) (see proposition 2.3.2).

By proposition 2.3.2 the group G acts transitively on the set B so by theorem 2.2.3there exists a covering ψ : U → P1(C)−∆ with characteristic subgroup GB1 , which hasindex m = N/n in G so the covering ψ has degree m.

(P1(C)− f−1(∆), t1)

f ))SSSSSSSSSSSSSSS(U, u0)

ψ

(P1(C)−∆, z0)

If H = Gt1 is the stabilizer of the element t1 ∈ f−1(z0) then H < GB1

∼= π1(U, u0),since H = Gt1 is the stabilizer in G of t1 ∈ B1, which is a block; so by theorem2.2.2 there exists a covering ϕ : (W,w0) → (U, u0) of degree n = [GB1 : H] withcharacteristic subgroup H. By proposition 2.2.1 the topological space W is isomorphicto P1(C)t−f

−1(∆) since their coverings over U have the same characteristic subgroup H;so without loss of generality we may assume that W = P1(C)t − f

−1(∆). The situationis the following

(P1(C)− f−1(∆), t1)ϕ

//

f ))SSSSSSSSSSSSSSS(U, u0)

ψ

(P1(C)−∆, z0)

By an argument similar to proposition 2.2.15 we can ”compactify” all the Riemannsurfaces to obtain finite ramified coverings of compact Riemann surfaces

(P1(C), t1)ϕ

//

f ''NNNNNNNNNNN(C, u0)

ψ

(P1(C), z0)


where C is a compact Riemann surface. By Riemann-Hurwitz theorem (for example)we have C ∼= P1(C) and so ϕ and ψ are rational functions.

Finally we have

f(t) = z = ψ(u) = ψ(ϕ(t))

So the function f is decomposable.

¿From previous theorem we deduce the following important corollary.

Corollary 2.3.4 Let f ∈ C(T ) be a rational function of degree N and let z0 ∈ P1(C) bea regular value of the covering map P1(C)→ P1(C), z 7→ f(z). Then there is a bijectionfrom the set of right components of decomposition of f modulo equivalence and the setof fundamental system of blocks of Mon(f) (which act over f−1(z0)), that is:

ϕ|f = ψ ϕ/ ∼↔ B = B1, . . . , Bm fundamental system of blocks for Mon(f)

The degree of ϕ corresponds to the cardinality of the corresponding block B. Moreoverϕ is indecomposable if and only if the associated block Bϕ cannot be decomposed insmaller blocks.

Proof : Let f = ψ ϕ and z0 be a regular value of f ; for each u ∈ ψ−1(z0), we setBu = t1, . . . , tn ∈ f

−1(v0)| ϕ(ti) = u. We have already seen that Bu is a Mon(f)-blockand the set B = Bu|u ∈ ψ

−1(z0) is a fundamental system of blocks for Mon(f) (seecorollary 2.2.7).

¿From the proof of the previous theorem we see that the map from the set of rightcomponents of f to the set of fundamental system of blocks of Mon(f), which sends ϕto Bϕ, is surjective.

We need only to prove that this map is injective, that is if f = ψ ϕ = ψ1 ϕ1 suchthat B = Bϕ = Bϕ then ϕ = µ ϕ1, for some linear rational function µ. Observe that inparticular we have deg(ϕ) = deg(ϕ1) = #B, where B ∈ B, and deg(ψ) = deg(ψ1) = #B.

We have the following diagram of ramified coverings

C

g

P1(C)

f

ϕ1

##HHH

HHHH

HHϕ

vvvv

vvvv

v

P1(C)

ψ ##HHH

HHHH

HHP1(C)

ψ1vvvv

vvvv

v

P1(C)

2.3. BLOCKS 37

where C is the Galois closure of f : P1(C)→ P1(C),

In terms of unramified coverings we have

(C, c0)

g

(W,w0)

f

ϕ1

%%KKKKKKKKKKϕ

yyttttttttt

(U, u0)

ψ %%JJJJJJJJJ

(U ′, u′0)

ψ1yyssssssssss

(V, v0)

This become the following diagram of inclusion of fields (see theorem 2.2.17)

L

C(T )

NNNNNNNNNNN

qqqqqqqqqq

C(ϕ)

MMMMMMMMMMC(ϕ1)

pppppppppp

C(Z) = C(f)

By corollary 2.2.18 we have Mon(ϕ g) ∼= J/K and Mon(ϕ1 g) ∼= J1/K, whereJ = π1(U, u0), J1 = π1(U

′, u′0) and K = π1(C, c0) = ker(Φ), where Φ is the monodromyhomomorphisms associated to f .

Observe now that by corollary 2.2.7 we have

J ∼= GBu0

and similarly

J1∼= GBu′

0

But since Bu0 and Bu′0 belong to the same fundamental system of blocks B thenGBu0

is conjugate to GBu′

0. Along with theorem 2.2.17 this implies that C(ϕ) ∼= C(ϕ1),

so ϕ is a linear rational function of ϕ1.


2.4 First theorem of Ritt

By lemma 2.1.6 a decomposition of a polynomial F in rational functions is equivalentto a decomposition in polynomials. This result is based on the fact that the mapF : P1(C) → P1(C) is totally ramified over ∞, in fact F−1(∞) = ∞; we get thesame result if F is totally ramified over a generic point of P1(C). Ritt’s results ondecomposition of polynomials are based on this simple observation.

Theorem 2.4.1 Let f ∈ C[T ] be of degree n. Then for each divisor d of n there existsat most one block of the monodromy group Mon(f) of cardinality d.

Proof : By proposition 2.2.10 the permutation σ∞ associated to the pathγ∞ ∈ π1(P

1(C) − ∆, z0) which turns around to ∞ is a n-cycle, since ∞ is a totallyramified point of the covering f : P1(C) → P1(C); let us suppose that σ∞ = (1, . . . , n)(by numbering the roots of f(T )− Z over C(Z) in a suitable way).

Let H∞ be the cyclic subgroup of G = Mon(f) generated by σ∞; a G-block is also aH∞-block, so by lemma 2.3.1 a G-block of cardinality d (n = dk) has the form

B = 1, k + 1, . . . , (d− 1)k + 1

and this proves the theorem.

¿From this result and from corollary 2.3.4 it follows that if F ∈ C[T ] of degree Nthen for each divisor d of N there exists at most one field K of degree d over C(F ) suchthat C(F ) ⊂ K ⊂ C(T ).

¿From now on, if f ∈ C[T ], we denote H∞ the cyclic subgroup of G = Mon(f)generated by the permutation σ∞ = (1, . . . , n). Obviously this subgroup acts transitivelyon the set of roots of f(T )− Z ∈ C(Z)[T ].

The following theorem (reformulation of lemma 2.1.9) tell us something more of theprevious theorem.

Theorem 2.4.2 Let f ∈ C[T ] and let ϕ,ψ ∈ C[T ] be indecomposable of degree n and mrespectively such that f = φ ϕ = σ ψ, for some φ, σ ∈ C[T ]. Then either n = m or(n,m) = 1.

Proof : Let N be the degree of f .It follows immediately from the previous theorem that the first case holds if and

only if ϕ and ψ are linearly equivalent (in fact if n = m then by corollary 2.3.4 we have#Bϕ = #Bψ and so by the previous theorem these fundamental system of blocks are thesame).

Suppose that n 6= m and let δ + (n,m) > 1; let N = nk = mh. Let B = Bϕ be thefundamental system of k blocks determined by ϕ:

B1 = 1, k + 1, . . . , (n− 1)k + 1, . . . , Bk = k, 2k, . . . , nk

2.4. FIRST THEOREM OF RITT 39

and let C = Bψ be the fundamental system of h blocks determined by ψ:

C1 = 1, h+ 1, . . . , (m− 1)h+ 1, . . . , Ch = h, 2h, . . . ,mh

By lemma 2.3.1 there exists a fundamental system of blocks D1, . . . , DN/δ ofH∞ =< σ∞ >, each of them of δ elements.

Since B1 is a G-block then it is also a H∞-block; by lemma 2.3.1 it follows thatD1 ⊂ B1. For the same reason D1 ⊂ C1.

Since the intersection of two G-blocks is a G-block then B1 ∩C1 is a G-block, whichis non-trivial because it contains D1 which has cardinality δ > 1. Then by corollary2.3.4 the polynomials ϕ and ψ would be decomposable, contrary to our assumption.

¿From this theorem it follows that if f is a polynomial such that f = φ ϕ = σ ψ,where ϕ and ψ are indecomposable of different degree (hence coprime degree), each blockof Bϕ has only one element in common with each block of Bψ, that is for each B ∈ Bϕand for each C ∈ Bψ we have #(B ∩ C) = 1.

We can now prove the first theorem of Ritt by induction on the degree of polynomialf . Up to degree 6 the theorem is true; we suppose the theorem for those polynomials fwith deg(f) < N , N > 6, and we prove it for those of degree N .

Let

f(X) = f1 . . . fr(X)

f(X) = g1 . . . gs(X)

be two maximal decomposition of a polynomial f of degree N ; if deg(fr) = deg(gs)then the two polynomials fr and gs determine the same fundamental system of blocks,so C(fr) = C(gs). By induction it follows that r − 1 = s − 1 and the degrees of thecomponents of the two decompositions are the same up to the order.

If deg(fr) = n 6= deg(gs) = m then (n,m) = 1 by previous theorem; then N isdivisible by nm and there exists a fundamental system of blocks D1, . . . , DN/nm of H∞,each of them of cardinality nm.

Lemma 2.4.3 The H∞-block D1 is a G-block.

Proof : We keep the notation of theorem 2.4.2.

For what we saw before D1 contains B1; moreover it follows from lemma 2.3.1 that

D1 =⋃

g∈HD1/HB1

g(B1)

where HD1 and HB1 are the stabilizers of D1 and B1 respectively in H∞ (that is theintersections of GD1 and GB1 respectively with H∞).

In the same way we have


D1 =⋃

g∈HD1/HC1

g(C1)

Then D1 is both the union of m blocks of B and the union of n blocks of C:

D1 =⋃

i∈I

Bi =⋃

j∈J

Cj

where #I = m and #J = n.¿From this identity and the fact that two blocks of B and C can have at most one

element in common (see previous theorem: gs and fr are indecomposable) it follows thatfor each i ∈ I and j ∈ J we have #(Bi ∩ Cj) = 1.

Set K + g ∈ G|g(1) ∈ D1. We have the following lemma.

Lemma 2.4.4 For each g ∈ K we have g(D1) = D1.

Proof : Let g ∈ K and let Bt be a block of B such that g(1) ∈ Bt ⊂ D1. Theng(B1) = Bt.

We have the following equalities

B1 =⋃

j∈J

B1 ∩ Cj , Bt =⋃

j∈J

Bt ∩ Cj

g(B1) =⋃

j∈J

g(B1 ∩ Cj) =⋃

j∈J

g(B1) ∩ g(Cj) =⋃

j∈J

Bt ∩ g(Cj)

⇒ g(Cj |j ∈ J) = Cj |j ∈ J ⇒ g(D1) = D1

which proves the lemma.

In particular from this lemma it follows that K is a group, and it is equal to thestabilizer of D1 in G: K = g ∈ G|g(D1) = D1.

Let g ∈ G be such that g(D1)∩D1 6= ∅: then there exist i, k ∈ I such that g(Bi) = Bk.

Bi =⋃

j∈J

Bi ∩ Cj , Bk =⋃

j∈J

Bk ∩ Cj

g(Bi) =⋃

j∈J

g(Bi ∩ Cj) =⋃

j∈J

g(Bi) ∩ g(Cj) =⋃

j∈J

Bk ∩ g(Cj)

⇒ g(Cj |j ∈ J) = Cj |j ∈ J ⇒ g(D1) = D1

Hence D1 is a G-block of cardinality nm. Lemma 2.4.3 is now proved.

Since D1 is a G-block of cardinality nm it follows by corollary 2.3.4 that there existsa polynomial µ ∈ C[T ] of degree nm such that f = σµ. Moreover we have µ = γfr andµ = ηgs, since D1 is both union of blocks of B and union of blocks of C.

2.4. FIRST THEOREM OF RITT 41

We summarize the situation in the following diagram:

C(T )

n

wwww

wwww

wm

GGGG

GGGG

G

nmC(fr)

m GGGG

GGGG

GC(gs)

nww

wwww

www

C(µ)

C(f)

where (n,m) = 1.So

f = f1 . . . fr = σ γ fr = σ η gs = g1 . . . gs

the following lemma concludes the proof of the theorem.

Lemma 2.4.5 The polynomials γ and η are indecomposables.

Proof : Like before we keep the notation of theorem 2.4.2.If γ = γ1γ2 with 1 < m2 = deg(γ2) < m then f = σµ = σγfr = σγ1(γ2fr);

by corollary 2.3.4 there exists a G-block E1 of cardinality nm2 such that B1 ⊂ E1 ⊂ D1.By lemma 2.3.1 we also have that (we recall that every G-block is also a H∞-block)

E1 =⋃Bi

Given a block Cj of C we have that

E1 ∩ Cj =⋃Bi ∩ Cj

and so the G-block E1 ∩Cj has cardinality m2 > 1; then gs would be decomposable,contradiction.

In the same way η is indecomposable.

By lemma 2.1.7 we have f1 . . . fr−1 = σ γ, so since this is a polynomial of degreeless than N , by induction we have that the number of indecomposable components of σis r − 2. For the same reason g1 . . . gs−1 = σ η, so the number of indecomposablecomponents of σ is s− 2. So r− 2 = s− 2 and the degrees of the components of the twodecompositions are the same up to the order.

If we have two maximal decompositions of a polynomial

f = f1 . . . fr = g1 . . . gr

is possible to pass from one to another through a finite number of steps:


• there exists i such that

f1 . . . fi fi+1 . . . fr = f1 . . . (fi λ) (λ−1fi+1) . . . fr

• there exists i such that fi fi+1 = gi gi+1 where deg(fi) = deg(gi+1) anddeg(fi+1) = deg(gi) such that

f1 . . . fi−1 (fi fi+1) fi+2 . . . fr = f1 . . . fi−1 (gi gi+1) fi+2 . . . fr

in this last case we have to apply Ritt’s second theorem.

Chapter 3

Plane algebraic curves

3.1 Affine curves

We adopt the definition of plane algebraic curve we are going to give; our base fieldis a fixed algebraic closure Q of Q. For a more general definition of algebraic curve(algebraic affine or projective variety of dimension one over an algebraically closed field)see other books like Shafarevich ([42]) or Hartshorne ([17]).

Definition 3.1.1 A plane affine algebraic curve (from now on simply curve) is asubset C of the affine plane A2(Q) defined as the zero-locus of a non-zero polynomial intwo variables f ∈ Q[X,Y ], that is:

C = Cf = (x, y) ∈ A2(Q)|f(x, y) = 0

The couples (x, y) ∈ C are called points of the curve.Let K be a number field. A curve C is defined over K if the polynomial defining

the curve is in K[X,Y ]. We indicate a curve C defined over a number field K as C/K.The points of C with coordinates in K are called K-rational points (or simply rationalif K = Q). We call C(K) the set of K-rational points.

A curve Cf defined over K is irreducible (resp. absolutely irreducible) if f(X,Y )is irreducible in the ring of polynomials K[X,Y ] (resp. irreducible in Q[X,Y ]).

If C/K is a curve, we define the coordinate ring of C as:

K[C] +K[X,Y ]

I(C/K)

where I(C/K) = f ∈ K[X,Y ]|f|C ≡ 0: we identify polynomials which coincideover C; observe that K[C] is an integral domain if and only if I(C/K) = (f) is a primeideal, which corresponds to the fact that f is irreducible in K[X,Y ]. In similar way wedefine Q[C].

In the case of an irreducible (resp. absolutely irreducible) curve, the field of quotientsof K[C] (resp. of Q[C]) is denoted by K(C) (resp. Q(C)) and it is called the field ofrational functions of the curve C with coefficients in K (resp. in Q).

43

44 CHAPTER 3. PLANE ALGEBRAIC CURVES

Note that the elements of Q[C] induce well-defined functions on C with values inQ which are called regular functions of the curve; the elements of the field Q(C) definefunctions on C which are regular over an open set (in the sense of Zariski topology, see[42]). We obviously have that K[C] = K[x, y] where x and y are the class modulo theideal I(C/K) of the coordinates X and Y , considered as regular functions on the curve.

The following classical theorem (see [42]) shows in effect that the ideal of an irre-ducible curve is generated by the polynomial f which defines the curve itself. If f is notirreducible it is generated by the radical of the ideal generated by the polynomial f .

Proposition 3.1.2 Let k be a field and let f ∈ k[X,Y ] be an irreducible polynomial. Ifg ∈ k[X,Y ] is not divisible by f then the sistem of equation f(x, y) = g(x, y) = 0 hasonly a finite number of solutions.

Lemma 3.1.3 An irreducible curve C defined over a number field K is absolutely ir-reducible if and only if the field K is algebraically closed in the field K(C) of rationalfunctions of the curve C with coefficients in K, that is K(C) ∩K = K, where K is afixed algebraic closure of the field K.

Proof : Let C = Cf , where f ∈ K[X,Y ] is irreducible and let K(C) = K(x, y).If f(X,Y ) is absolutely irreducible let K(C) = K(x, y). It follows that

[K(C) : K(x)] = [K(C) : K(x)]

By corollary 9.2 of [14] it follows that K(C) and K(x) are linearly disjoint over K(x).Since K and K(x) are linearly disjoint over K it follows that K and K(C) are linearlydisjoint over K which in particular implies that K(C) ∩K = K.

Conversely if K is algebraically closed in K(C) then K and K(C) are linearly disjointover K (see theorem 2 pg. 56 Lang, [27]) and therefore for the tower property ([26],Prop.1 pg. 50) it follows that K(x) and K(C) are linearly disjoint over K(x) and forthe previously mentioned corollary of [14] we have that [K(C) : K(x)] = [K(C) : K(x)],which implies that f(X,Y ) is irreducible in Q[X,Y ]. The following diagram summarizesthe situation:

K(x, y)

vvvv

vvvv

v

JJJJJJJJJJ

K(x)

IIIIIIIII

K(x, y)

ssssssssss

K

EEEE

EEEE

E K(x)

tttttttttt

K

3.2. PROJECTIVE CURVES 45

Every algebraic curve is defined over a number field: take the finite extension of Qgenerated by the coefficients of the polynomial f(X,Y ) ∈ Q[X,Y ] which defines thecurve.

Observe also that a curve C can be defined over a number field K but its set ofK-rational points can be empty, see for example the curve defined by the absolutelyirreducible polynomial f(X,Y ) = X2 + Y 2 + 1 which has no rational points.

The set C(Q) is always infinite (see [20]).

Definition 3.1.4 A point p ∈ Cf is non-singular if at least one of the two derivativesof f in p is non-zero. A curve C is non-singular if all of its points are non-singular.

The following proposition holds (see [42], chap. 2, §. 5) :

Proposition 3.1.5 An absolutely irreducible curve C is non singular if and only if itscoordinate ring Q[C] is integrally closed.

If an absolutely irreducible curve C/K is non singular then also K[C] is integrallyclosed.

An absolutely irreducible curve C is a rational curve (also said of genus zero sincethe desingularization of C is a compact Riemann surface of genus zero) if its points canbe parametrized by rational functions of a parameter t, that is there exist two rationalfunctions ϕ,ψ ∈ Q(t) such that for almost every points p = (x, y) of the curve (except afinite number of points) there exists t ∈ Q such that (x, y) = (ϕ(t), ψ(t)).

The parametrization (ϕ,ψ) is proper if for almost every point p = (x, y) of the curvethere exists exactly one t ∈ Q such that (x, y) = (ϕ(t), ψ(t)); observe that a parametriza-tion is proper if and only if the couple of rational functions of the parametrization isminimal (see paragraph 1.2). By Luroth’s theorem every rational curve admits a properparametrization.

If the components of the parametrization ϕ,ψ can be chosen in Q[t] then we saythat the curve is polynomially parametrized or the curve is polynomial. The polynomialversion of Luroth’s theorem (see theorem 1.1.4) shows that if a rational curve admits apolynomial parametrization then it admits a proper polynomial parametrization.

3.2 Projective curves

Let P2(Q) + (Q3−(0, 0, 0))/∼ be the projective plane over Q with usual equivalence

relation: given x = (x0, x1, x2) and y = (y0, y1, y2) in Q3− (0, 0, 0) we define x ∼ y

if and only if there exists λ ∈ Q∗

such that λx = y. We denote x = (x0 : x1 : x2)the equivalence class of x = (x0, x1, x2) and P2 the projective plane P2(Q). We remindthat P2 can be covered by subsets which are copies of A2, for example Ui + (x0 :x1 : x2)| xi 6= 0, for i = 0, 1, 2; in particular A2 is a subset of P2, through the map(x : y) 7→ (x : y : 1) (if i = 1; likewise if i = 2 or 3); the image points of A2 through this


map are called finite points of P2 and the points (x0 : x1 : 0) are called points at infinityof P2.

Definition 3.2.1 A plane projective curve C is a subset of the projective plane P2

defined as the zero-locus of a non-zero homogeneous polynomial f ∈ Q[X,Y, Z]:

C = Cf = (x : y : z) ∈ P2| f(x : y : z) = 0

In the same way of affine curves we define points of projective cruves, projectivecurves defined over a number field and (absolutely) irreducible projective curves.

Every affine curve can be considered as a subset of a projective curve by homogenizingthe polynomial which defines the curve. If Cf ⊂ A2 is an affine curve, we define itsprojective closure as the projective curve Cf given by the homogeneous polynomial

f associated to f :

f(X,Y, Z) + Zdeg(f)f(X/Z, Y/Z)

Conversely if Cf ⊂ P2 is a projective curve we define its dehomogeneization with

respect to Z as the curve Cf ⊂ A2 given as the zero-locus of the polynomial f(X,Y ) =f(X,Y, 1) (this in the case when Z does not divide f). From a geometric point of view,the projective curve Cf is the topological closure of Cf in P2 with respect to the Zariski

topology (if we see Cf ⊂ A2 in P2 through the immersion A2 → P2 defined above; see[42]).

If C is a curve we call points at infinity of C the elements of C−C: they correspondto the points of P1 which satisfies fd(X0, X1) = 0, where fd is the homogeneous part ofhighest degree of the polynomial f which defines the curve C. So the number of pointsat infinity of an affine curve are equal to the number of irreducible distinct factor of thehomogeneous component of f of highest degree. Obviously we have C = C ∩ A2.

A point p ∈ C is non singular if there exists i ∈ 1, 2, 3 such that p ∈ Ui and p isnon singular for the affine curve Ui ∩ C.

We have to pay more attention to the definition of rational function field of a irre-ducible projective curve C. Let f, g ∈ Q[X,Y, Z] be two homogeneous polynomials of thesame degree such that g 6∈ I(C): then f/g is a well-defined function in the points of Cwhere g 6= 0. The set of all these functions is a ring OC ; the quotient of OC for the max-imal ideal MC of the funtions f/g ∈ OC such that f ∈ I(C) is the field Q(C) of rationalfunctions of C (we identify two rational functions f/g and f ′/g′ if g′f − gf ′ ∈ I(C)).

¿From now on we assume that our curves, affine or projective, are absolutely irre-ducible.

3.3. RATIONAL MAPS BETWEEN CURVES 47

3.3 Rational maps between curves

Now we define map between curves; first we treat the case of affine curves, then thecase of projective curves.

• Affine case

Definition 3.3.1 Let C1 and C2 be curves. A rational map from C1 to C2 is a map

φ : C1 → C2

such that there exist φ1, φ2 ∈ Q(C1) such that for each p ∈ C1 where φ1, φ2 are bothdefined we have φ(p) = (φ1(p), φ2(p)) ∈ C2.

If K is a number field and C1, C2 are defined over K, the map φ is defined over Kif φ1, φ2 ∈ K(C1).

A rational map φ : C1 → C2 between two curves is defined over an open set of C1

(with respect to the Zariski topology of A2(Q)), that is in the points of C1 where φ1 andφ2 are defined. If φ is defined over a number field K then it restricts to a map from theK-rational points of C1 to the K-rational points of C2, that is φ : C1(K)→ C2(K).

• Projective case

Definition 3.3.2 Let C1 and C2 be projective curves. A rational map from C1 to C2

is a map

φ : C2 → C2

of the form φ = [ϕ1 : ϕ2 : ϕ3] where ϕi ∈ Q(C1) for i = 1, 2, 3 and for each P ∈ C1

where ϕi are defined we have φ(P ) = [ϕ1(P ) : ϕ2(P ) : ϕ3(P )] ∈ C2.If K is a number field and C1, C2 are defined over K, the map φ is defined over K

if there exists λ ∈ Q such that λϕ1, λϕ2, λϕ3 ∈ K(C1) (note that φ = [λϕ1 : λϕ2 : λϕ3]).

If φ = [ϕ1 : ϕ2 : ϕ3] is a rational map between two projective curves with ϕi =φi/ψi ∈ Q(C1) and φi, ψi are homgeneous polynomials of the same degree ni, we considerthe homogeneous polynomial ψ = mcmψi and we multiply the three rational functionsϕi by ψ. We obtain the following expression for φ

φ = [f1 : f2 : f3]

where fi are homogeneous polynomial of the same degree. If there is some commonfactor between the fi we can divide by it in order to suppose that the fi’s are coprime.

Let φ : C1 → C2 be a rational map between two irreducible affine curves, whereCi = (x, y) ∈ A2|fi(x, y) = 0; suppose also that the projective closure of C1 is non-singular. We want to show that φ can be extended to a rational map φ : C1 → C2

between the projective closure of C1 and C2 respectively.


First of all if such an extension exists then it is unique by lemma 4.1, chap. 1 of [17],since by 3.3.6 the map φ is regular.

If φ = (φ1, φ2), where φi ∈ Q(C1) is equal to ϕi/ψi, we set

φi(X,Y, Z) = (ϕi(X/Z, Y/Z)Zdeg(φi))/(ψi(X/Z, Y/Z)Zdeg(φi)) = ϕi(X,Y, Z)/ψi(X,Y, Z)

This rational function is a ratio of two homogeneous polynomials of the same degreedeg(φi). We then define the map φ

φ = [φ1 : φ2 : 1] = [φ′1 : φ

′2 : ψ]

where ψ = mcmψ1, ψ2 and φ′1, φ

′2, ψ are homogeneous polynomials of the same

degree. Since

f2

(ϕ1

ψ1,ϕ2

ψ2

)= 0

we have that

F2(φ′1, φ

′2, ψ) = 0

If D ⊂ C1 is the domain of definition of φ (the complementary of the set of zeroof ψ1 and ψ2 in C1) then it is immediate to see that for each (x, y) ∈ D we haveφ(x, y) = φ(x : y : 1).

If (x, y) is a pole either of φ1 or of φ2 then

φ(x : y : 1) = [φ′1(x : y : 1) : φ

′2(x : y : 1) : 0] ∈ C2 − C2

so the points where φ is not defined are sent through φ to the points at infinity ofC2, that is the set C2 − C2.

C1 → C1

↓ ↓C2 → C2

It can happen that a point of C1 − C1 is sent through φ to a point of C2. Take forexample the map φ : A1 → C, where C = (x, y)|x2 + y2 = 1, given by

t 7→

(2t

t2 + 1,t2 − 1

t2 + 1

)

The corresponding map φ : P1 → C = (x : y : z)|x2 + y2 = z2 is given by

(t : s) 7→ (2ts : t2 − s2 : t2 + s2)

We have φ(1, 0) = (0 : 1 : 1).

3.3. RATIONAL MAPS BETWEEN CURVES 49

Proposition 3.3.3 A regular map between two curves φ : C1 → C2 is constant orsurjective.

Proof : The proof follows from the fact that φ is a finite map since C1 and C2 arealgebraic varieties of dimension 1 (see Shafarevich, [42], chap.1 §5.3).

Proposition 3.3.4 Let C1/K and C2/K be curves and let φ : C1 → C2 be a rationalmap defined over K, such that φ(C1) is dense in C2 (we call such a map dominant).Then the map

φ∗ : K(C2)→ K(C1)

ϕ 7→ ϕ φ is well defined and K(C2) ⊂ K(C1) is a finite extension of fields.

Proof : The map φ∗ : K(C2) → K(C1), ϕ 7→ ϕ φ, is well defined since ϕ|C2= 0

implies ϕ φ|C1= 0. It is also injective: if f ∈ K[C2] such that φ∗(f) = 0 it follows that

f|φ(C1) = 0 , so φ(C1) ⊂ f(p) = 0 which means C2 = φ(C1) ⊂ f(p) = 0.This map extends in a natural way to a map φ∗ : K(C2) → K(C1) which is an

inclusion of field; this extension is finite since K(C2) and K(C1) have trascendencedegree 1 over K.

Definition 3.3.5 If φ : C1 → C2 is a dominant rational map defined over a numberfield K between two curves C1/K,C2/K, then the degree of the map φ is defined as thedegree of the extension of fields K(C2) ⊂ K(C1).

Proposition 3.3.6 Let φ : C1 → C2 be a rational map between two projective curves;if p ∈ C1 is a non-singular point then φ is defined in p.

For a proof of this simple fact, which uses only that Op, the local ring of C1 at p, isa discrete valuation ring, see Proposition 2.1 of [41].

Definition 3.3.7 A non-constant rational map φ : C1 → C2, where C1, C2 are curves,is called birational if there exists a rational map ψ : C2 → C1 sich that ψ φ = IdC1 eφ ψ = IdC2 (on the open set where this functions are defined).

We can say that a rational curve is a curve which is birational to A1 (or to P1, if thecurve is projective).

In general a curve C is birational to a curve C ′ if and only if their rational functionfields are isomorphic. More precisely there is an equivalence between the category ofabsolutely irreducible curve defined over a number field K with morphisms given bydominant rational map defined over K which are not constant and the category ofalgebraic function fields in one variable K/K with K ∩Q = K and morphisms given byinclusion of fields which leaves K fixed; this functor associates to a curve C/K the fieldK(C) and to a dominant rational map φ : C1 → C2 defined over K the inclusion of fieldsφ∗ : K(C2)→ K(C1), which restricted to K is the identity.


3.4 Geometric points and places

Now we show that geometric points of a non-singular curve correspond bijectively tovaluation rings of the field Q(C) which contain the base field Q; the latter are usuallycalled places of the curve.

Lemma 3.4.1 Let C be an absolutely irreducible, non-singular curve defined over anumber field K and let p be a point of C; let Op,K = f ∈ K(C)|f is defined in p thelocal ring of C in p with coefficients in K andMp,K = f ∈ Op,K |f(p) = 0 its maximalideal.

If p = (a, b) then the residue field K(p) + Op,K/Mp,K is equal to K(a, b) and so itis a finite extension of K.

Proof : Since K ∩ Mp,K = 0 then K ⊂ K(p). Consider the projection mapπ : Op,K → Op,K/Mp,K ; the image set of π is equal to f(x, y)|f ∈ Op,K, wherex = π(x) and y = π(y) (x and y are elements of Op,K).

Take now the minimal polynomials g(T ) and h(T ) over K of a and b respectively;the non-zero regular functions g(x) and h(y) are obviously contained Mp,K , so theirimages g(x) and h(y) are zero in K(p). Hence x and y are algebraic over K and theysatisfy the same minimal polynomial of a and b respectively over K. So the field K(p)is isomorphic to the finite algebraic extension K(a, b).

A simple consequence of this lemma is the fact that a point p of a curve C/K isdefined over K (that is an elements of C(K)) if and only if its residue field is equal toK. The residue field K(p) is also called the field of definition of the point p.

Proposition 3.4.2 Let C be a non-singular, projective and absolutely irreducible curve.Then the points of the curve are in bijection with the valuation rings of the field Q(C)which contains the base field Q.

Moreover if C is defined over a number field K, its rational points C(K) are inbijection with the valuation rings of K(C) whose residue field is isomorphic to K.

Proof : Let p be a point of C, an affine model of C. We associate to p the localring of C in p, that is Op = f ∈ Q(C)|f is regular in p, which is the localized ofQ[C] at the maximal ideal of the regular functions of Q[C] which are zero in p. It isa discrete valuation ring since the curve has dimension one and p is non singular (seeShafarevich, [42]). This map is injective since there is a bijection between the points ofC and maximal ideals of Q[C] (by Hilbert’s Nullstellensatz theorem).

Conversely let O be a valuation ring of Q(C) = Q(x, y) which contains Q(x); thenA = O ∩ Q(x) is a valuation ring of Q(x) which contains x and it corresponds to anelement x0 ∈ Q (see proposition 1.3.4). Hence x − x0 is a generator for the maximalideal P = Px0 of A = Ax0 ; since by lemma 1.3.5 we can assume that y is integral overQ[x] then y ∈ O by proposition 1.3.6. So there exists y0 ∈ Q such that y − y0 ∈M , the

3.5. PARAMETRIZATION WITH RATIONAL COEFFICIENTS 51

maximal ideal of O; then the relation f(x, y) = 0 in Q(C) implies f(x0, y0) = 0 in Q,and so p0 = (x0, y0) ∈ Cf is the associated point of the valuation ring O.

The point p0 is unique, since it is uniquely determined by the maximal idealM ∩Q[x, y].

If x 6∈ O then it is easy to observe that there exists another affine model C ′ of Csuch that Q[C ′] = Q[x′, y′] ⊂ O.

Let C be an algebraic curve; if S is the set of valuation rings of Q(C) then wecan give a topology to S in order to obtain a compact Riemann surfaces which is thedesingularization of the curve C: there exists a holomorphic map π : S → C such thatπ|S−π−1(∆) : S−π−1(∆)→ C−∆ is biholomorphic, where ∆ is the set of singular pointsof C.

For more references see chapter 2 of [26] and also [7].

3.5 Parametrization with rational coefficients

If a rational curve is defined over a number field K it is not true in general thatit admits a parametrization with coefficients in the field K; for example the curvef(X,Y ) = X2 + Y 2 + 1 which is rational but Cf (Q) = ∅.

If the curve is non-singular and the set of rational points is not empty then thereexists a parametrization with rational coefficients, as the following proposition shows.

Proposition 3.5.1 Let C be a non-singular rational curve, defined over Q and abso-lutely irreducible. If C has at least one rational point, then there exists a parametrizationof C which is defined over Q, that is a couple of rational functions (ϕ,ψ) with coefficientsin Q such that f(ϕ(t), ψ(t)) = 0.

Proof : Let p ∈ C(Q) and consider the set

L(p) = ϕ ∈ Q(C)|ϕ has at most a simple pole in p

This is a vector space over Q and since the curve is rational then by Riemann-Rochtheorem the space L(p) has dimension 2 over Q (see [18]). Its elements are the constantsand functions which have a simple pole in p ; the latter correspond to rational mapsfrom C to A1(Q) of degree 1.

We want to show that there exists a base of L(p) of the form 1, ϕ, where 1 is theconstant function and ϕ ∈ Q(C) − Q. First we observe that L(p) is stable under theaction of the Galois group G = Gal(Q/Q), since p has coordinate in Q: if ϕ ∈ L(p) isnot constant and σ ∈ G then σ(ϕ) has a simple pole in ϕ(p) = p.

If φ ∈ L(p) then there exists a number field K such that φ ∈ K(C); without loss ofgenerality we can suppose that K/Q is a Galois extension. Let α1, . . . , αn a base of Kover Q and let Gal(K/Q) = σ1, . . . , σn; it is well known that the determinant of thematrix M = (σi(αj))1≤i,j≤n is different from zero (see Lang, [24]).


Now we define the following rational functions, for j = 1, . . . , n :

gj +∑

i=1,...,n

σi(αjφ) =∑

i=1,...,n

σi(αj)σi(φ)

It is immediate to verify that gj are invariant under the action of Gal(K/Q), sothey belong to the field Q(C); since the matrix M with coefficients in K is invertiblethen it is possible to express σi(φ) (and in particular φ) as a linear combination withcoefficients in K of g1, . . . , gn ∈ Q(C). Hence every elements of L(p) is a Q-linearcombination of elements of Q(C). Let ϕ ∈ Q(C)−Q be such that 1, ϕ is a base of L(p)as a Q-vector space.

To conclude the proof we observe that the rational function ϕ ∈ Q(C) ⊂ Q(C)determines a rational map from C to A1(Q) of degree 1 which is defined over Q; thismap is an isomorphism since C is non-singular (see 3.3.3). Its inverse ψ : A1(Q)→ C isdefined over Q and it has the form ψ(t) = (ψ1(t), ψ2(t)), ψ1, ψ2 ∈ Q(t).

¿From the remarks of the previous paragraphs, in the same hypothesis of the propo-sition we can find a parametrization (ϕ(t), ψ(t)) with rational coefficients which is alsoproper.

More in general if a rational curve has a rational point which is non singular thenit has a parametrization with rational coefficients. Similar arguments if we replace thefield Q with a number field K.

So in effect almost every rational points of the curve can be obtained in terms of aparametrization with rational coefficients by choosing a rational parameter t.

3.6 Standard parametrization of rational curves

An example of rational curve defined over Q with a parametrization with rationalcoefficients is given by f(X,Y ) = X2 + Y 2 − 1, which is parametrized by the rationalfunctions (ϕ(t), ψ(t)) = (2t/(t2 + 1), (t2 − 1)/(t2 + 1)).

This parametrization of the curve Cf has a particular form: each component has thesame denominator (that is they have the same set of poles). This can be done in generalup to birationality.

Proposition 3.6.1 Let C be an irreducible curve defined over Q; then there exists amodel C ′ birational to C such that if Q(C ′) = Q(x′, y′) then x′ and y′ have the samedivisor of poles.

Proof : The statement means that x′ and y′ have the same set of poles with thesame multiciplicity. Suppose that Q(C) = Q(x, y) and consider the rational lineartransformations Φ with coefficients a, b, c, d in Q defined on Q(C):

Φ(x, y) + (x′, y′) = (ax+ by, cx+ dy)

3.6. STANDARD PARAMETRIZATION OF RATIONAL CURVES 53

such that ad− bc 6= 0 and abcd 6= 0; its image is Q(x′, y′) = Q(C).This map establishes an isomorphism of field Q(x, y) ∼= Q(x′, y′) which corresponds toa birationality of C with a curve C ′, whose defining polynomial f ′(X,Y ) is given byf ′(x′, y′) = 0.

We want to show that there exist a, b, c, d ∈ Q such that the statement of the propo-sition holds.

We set Px the support of the divisor of poles of x and Py the support of the divisorof poles of y; let P = Px ∪ Py be the set of poles of x or y. Obviously if p ∈ (Px − Py) ∪(Py − Px) then p is a pole for x′ and y′ with the same multiciplicity; if p ∈ Px ∩ Py wedistinguish two cases:

• ordp(x) = ordp(y) = k

If t is a local parameter in p we write x and y as Laurent expansion in a neighbourhoodof p:

x =akt−k

+ . . .+ a0 + a1t+ . . .

y =bkt−k

+ . . .+ b0 + b1t+ . . .

Then ordp(x′) = ordp(ax + by) = k provided (a, b) 6∈ (u, v)|aku + bkv = 0. Same

arguments for y′.

• ordp(x) 6= ordp(y)

In this case ordp(x′) = ordp(ax + by) = minordp(x), ordp(y) = ordp(cx + dy) =

ordp(y′) .

If we choose the coefficients a, b, c, d of the map Φ outside a finite number of hyper-planes we obtain what we wanted.

Corollary 3.6.2 Let C be a rational non-singular curve which is absolutely irreducible.Then there exists a curve C ′ birational to C such that it admits a parametrization of thefollowing type:

(ϕ(t), ψ(t)) =

(ϕ1(t)

φ(t),ψ1(t)

φ(t)

)

where ϕ1, ψ1, φ ∈ Q[t].

Proof : Suppose that the curve C is defined over a number field K; by proposition3.6.1 there exists a model C ′ of the curve C such that K(C ′) = K(x, y) and the rationalfunctions x and y of the curve C ′ have the same divisor of poles.

By hypothesis K(C ′) is a purely trascendental field over K of degree one; so thereexists t ∈ K(C ′) such that K(x, y) = K(t) and x = ϕ(t), y = ψ(t), where ϕ,ψ ∈ K(T ).


Since x and y have the same divisor of poles, and these correspond to the poles of therational functions ϕ and ψ, we have that

ϕ(t) =φ1(t)

φ(t)

ψ(t) =ψ1(t)

φ(t)

The corollary is proved.

3.7 Polynomial parametrization of rational curves

The following theorem (see for example [1]) gives a criterium to determine whethera rational curve C has a polynomial parametrization.

Theorem 3.7.1 Let C be a non-singular rational projective curve and C = C ∩ U3 bean affine model of the curve. Then C admits a polynomial parametrization if and onlyif C has exactly one point at infinity.

Proof : Suppose that C has a polynomial parametrization Φ of the following type

Φ : A1 → C

t 7→ (ϕ(t), ψ(t))

where ϕ(t), ψ(t) ∈ Q[t]; if we compose this map with the canonical inclusion of A2

in P2, (x, y) 7→ (x : y : 1), we obtain the map t 7→ (ϕ(t) : ψ(t) : 1), from A1 to C. Byproposition 3.3.3 the map Φ is surjective since Φ is regular on A1.

A1 Φ // _

C _

P1 Φ // C

Consider the extension Φ of Φ at P1 as we saw in section 3.3; by proposition 3.3.6,Φ is regular at (1 : 0) ∈ P1 − A1 (which is the point at infinity) so Φ is surjective byproposition 3.3.3. This implies that C−C has only one point, which is given by Φ(1 : 0).

Conversely suppose that C − C = P0 and let Φ(t) = (ϕ(t)/φ1(t), ψ(t)/φ2(t)) be aparametrization of C; we have

Φ(t : s) = (ϕ′(t, s) : ψ′(t, s) : φ(t, s))

where ϕ′, ψ′, φ are homogeneous polynomials of the same degree n and Φ : P1 → Csuch that Φ|A1 = Φ. Since P1 and C are non-singular and the map Φ has degree 1, it isan isomorphism.

3.7. POLYNOMIAL PARAMETRIZATION OF RATIONAL CURVES 55

We have

Φ−1

(C − C) = (t : s) ∈ P1|φ(t, s) = 0

and by the previous remark Φ−1

(C − C) has only one point in P1. Hence the poly-nomial φ has the form

φ(t, s) = (αt− βs)n

for some α, β ∈ Q.

We have

Φ(t) = Φ(t : 1) = (ϕ′(t, 1) : ψ′(t, 1) : (αt− β)n)

and if (t : 1) 6= (β/α : 1) we have

Φ(t) =

(ϕ′(t)

(αt− β)n,

ψ′(t)

(αt− β)n

)

If we compose the map Φ(t) with the rational linear function λ(t) = 1/t+β/α (whichis a birationality of A1 with itself) we obtain a polynomial parametrization Φ λ(t) aswanted.

In general if C is a rational plane curve (singular or not) we have to check that C−Chas only one point P0 and also that the curve is analytically irreducible at P0 in orderto obtain a polynomial parametrization.

If P0 is non singular then there is only one place at P0; the converse is not true: takefor example the curve y2 = x3 which is singular at the origin but there is only one placeat the origin.

It is useful to rephrase the same result from a completely algebraic point of view.We have the following proposition (see [13]).

Proposition 3.7.2 Let k be a field and t a trascendental element over k; let z be arational function in t, that is an element of k(t).

Then z is a polynomial in t if and only if the valuation ring at infinity O∞ of k(t)is the only valuation ring of k(t) over the valuation ring at infinity o∞ of k(z).

If this happens then O∞ is totally ramified over o∞.

Proof : First we observe that if a valuation ring O with maximal ideal B of k(t) isabove O1/z then 1/z ∈ P , since 1/z is a local parameter of o∞; this means that z 6∈ O.Conversely if a valuation ring O of k(t) does not contain z then O ∩ k(z) = o∞.

Suppose that O∞ is the only valuation ring of k(t) above o∞; the integral closure ofk[t] in k(t) (which is equal to k[t] since it is integrally closed) is equal to the intersectionof all the valuation rings of k(t) which contain k[t] (see proposition 1.3.6), that is allthe valuation rings of k(t) with the only exception of O∞. But z belongs to all of them


because the only valuation ring of k(t) which does not contain z is O∞, by the previousremark.

Conversely let z = f(t) ∈ k[t]; suppose that Oa is a valuation ring of k(t) differentfrom O∞ which is above o∞. Then 1/z ∈ Ba, the maximal ideal of Oa, so 1/z =(t−a)nf(t)/g(t), with n ≥ 1 e g, f coprime, f(a)g(a) 6= 0; hence z = 1/(t−a)ng(t)/f(t)which does not belong to k[t], contradiction.

The valuation ring at infinity of k(t) has residue field O∞/B∞ = k; in the same wayfor the valuation ring at infinity of k(z). Hence if O∞ is the only valuation ring aboveo∞, it is totally ramified.

Proposition 3.7.3 Let C ⊂ A2 be a rational irreducible curve. Then C admits a poly-nomial parametrization if and only if there exists a unique valuation ring O of Q(C)above the valuation ring at infinity of Q(x).

Proof : As we have already seen we can suppose that y is integral above Q[x], sothe set of poles of y is contained in the set of poles of x. By hypothesis there exists t ∈ Qsuch that Q(C) = Q(t); then there exists a unique valuation ring O of Q(C) above thevaluation ring at infinity of Q(x) if and only if x has a unique pole P0 on C which ispole also for y (the only rational functions of a curve which does not have poles are theconstants).

With same reasonings of theorem 3.7.1 we can compose with an automorphism ofP1 ∼= C to bring P0 at infinity, so x and y are polynomials in a parameter t.

We can also say that the curve C has only one place at infinity if and only if thereis only one place of Q(C) which does not contain the coordinate ring Q[C] of the curve.

Of course if C is a rational projective curve, there can be affine models of C witha polynomial parametrization and other affine models which do not have a polynomialparametrization. For example the projective curve C = (X : Y : Z) ∈ P2|XY = Z2has the affine model Cz = (x, y) ∈ P2|xy = 1 which does not have a polynomialparametrization; instead the affine model Cy = (x, z) ∈ P2|x = z2 has a polynomialparametrization.

Chapter 4

Curves of Schinzel

The problem of this chapter is to characterize pairs (f, S), where f ∈ Q[X,Y ] is anirreducible polynomial and S is an infinite subset of Q such that for all x in S thereexists y in S such that f(x, y) = 0.

We shall see that in the case of rational curves this problem is related to theparametrization of image set of rational functions (or polynomials, if the curve hasa polynomial parametrization).

4.1 Introduction

We give the following definition:

Definition 4.1.1 A polynomial f(X,Y ) ∈ Q[X,Y ] has the Schinzel’s property (SP)if there exists an infinite subset S = Sf ⊂ Q such that for every x ∈ S the equationf(x, Y ) = 0 has a solution y ∈ S.

This lemma is straightforward.

Lemma 4.1.2 Let f ∈ Q[X,Y ] and suppose that f is symmetric (that is f(X,Y ) =f(Y,X) ) and Cf (Q) is infinite. Then f has the SP with S = p1(Cf (Q)), wherep1 : Cf (Q)→ A1(Q) is the projection on the first coordinate.

Let f be a polynomial with SP; from the property of the set S we can constructsequences of elements of S in the following way.

Let x0 ∈ S: there exists x1 ∈ S such that f(x0, x1) = 0. Since x1 ∈ S there existsx2 ∈ S such that f(x1, x2) = 0, and so on. If there exists k,m ∈ N, k < n such thatxk = xm then we say that xnn∈N is a preperiodic sequence; if k = 0 the we call itperiodic sequence. In the other case, if xnn∈N goes on without repetition, then we callit infinite sequence.

We can also recursively define the set S as :

57

58 CHAPTER 4. CURVES OF SCHINZEL

S0 = x0 ∈ Q |∃x1 ∈ Q s.t.f(x0, x1) = 0 = p1(Cf (Q))

S1 = x0 ∈ S0 |∃x2 ∈ Q s.t.f(x1, x2) = 0 = x0 ∈ S0|x1 ∈ S0

. . .

Sn = x0 ∈ Sn−1 |∃xn+1 ∈ Q s.t.f(xn, xn+1) = 0 = x0 ∈ S0|x1 ∈ Sn−1

Note that for each n ∈ N we have Sn ⊂ Sn−1.

We can set

S =⋂

n=0,...,∞

Sn (4.1)

Now we use the following theorem of Faltings, which proved a conjecture by Mordell(see [18]):

Theorem 4.1.3 (Faltings) Let C an irreducible curve defined over Q. If C(Q) isinfinite then the genus of C is 0 or 1.

We cite also the following theorem due to Siegel which deal with the case of curveswith infinite integer points (see [39]).

Theorem 4.1.4 (Siegel) Let C be an irreducible curve defined over Q. If C(Z) isinfinite then the genus of C is 0 and the desingularization C of C has at most two pointsat infinity.

By Falting’s theorem if f(X,Y ) is an irreducible polynomial with Schinzel’s propertythen its associated curve has genus zero or one. In this chapter we consider only thecase of genus zero curves.

If f(X,Y ) defines a rational curve, by definition there exist rational functions ϕ,ψ ∈Q(t) such that

f(ϕ(t), ψ(t)) = 0

We may assume that

• ϕ,ψ ∈ Q(t) since Cf (Q) is not empty and there exists a non singular point ofthe curve in Cf (Q) since there are at most a finite number of singular points (seesection 3.5)

• the couple (ϕ,ψ) is minimal (see the definition in section 1.2)

4.1. INTRODUCTION 59

We impose also that ϕ(t), ψ(t) are polynomials, which it is equivalent to say that thedesingularization of the curve C has only one point at infinity (see section 3.7).

In the case of rational curves whose polynomial has Schinzel’s property we reducethe problem to the study of parametrization of the set of rational values of a rationalfunction with another rational function.

More precisely, let (ϕ(t), ψ(t)) be a fixed parametrization of a rational curveC = (x, y) ∈ A2|f(x, y) = 0; then there exists an infinite subset S′ of the rationalnumbers such that

S = ϕ(t)|t ∈ S′

since S is a subset of the projection on the first coordinate of the set of rationalpoints of the curve, which is equal to ϕ(t)|t ∈ Q.

By an argument similar to the one of the previous paragraph, for each t0 ∈ S′

the point (ϕ(t0), ψ(t0)) is in C, with ψ(t0) ∈ S. So there exists t1 ∈ S′ such thatψ(t0) = ϕ(t1); since t1 ∈ S

′ there exists t2 ∈ S′ such that ψ(t1) = ϕ(t2) and so on.

Hence if (ϕ(t), ψ(t)) is a parametrization of a rational curve with Schinzel’s property,we have that

ψ(S′) ⊂ ϕ(S′)

for some S′ infinite subset of the rational numbers.

We conjecture that there exists r ∈ Q[t] such that

ϕ(r(t)) = ψ(t)

So we have the following:

Conjecture 1

Let (ϕ(t), ψ(t)) a minimal couple of rational functions with rational coefficients andlet S an infinite subset of the rational numbers such that

ψ(S) ⊂ ϕ(S)

Then there exists r ∈ Q(t) such that ϕ(r(t)) = ψ(t).

Of course conjecture 1 is true when S = Q or also S = Z (for example by Hilbert’sirreducibility theorem). Moreover if in these cases we have that ψ(S) = ϕ(S) then thereexists a linear r ∈ Q(t) such that ϕ(r(t)) = ψ(t).

Conjecture 1 implies this other one:

Conjecture 2

Let f(X,Y ) an irreducible polynomial with rational coefficients and with Schinzel’sproperty. Let S ⊂ Q be the set as defined in (4.1).


If the curve C = (x, y) ∈ A2|f(x, y) = 0 is rational then there exists a parametriza-tion of C of the form

f(ϕ(t), ϕ(r(t))) = 0

where ϕ and r are rational functions, and the set S is equal to ϕ(t)|t ∈ Q.

If we do not suppose that the couple (ϕ(t), ψ(t)) is minimal then our first conjectureis false. For example take the polynomials

ϕ(t) = t2

ψ(t) = −t2 + 1

This couple of polynomials is not minimal since [Q(t) : Q(ϕ,ψ)] = 2. It can beimmediately seen that there is no rational function r such that ψ(t) = ϕ(r(t)) (note thatr should have degree one).

If we set S = π1(Cf (Q)), where f(X,Y ) = X2 + Y 2 − 1, we see that ψ(S) = ϕ(S).In particular the polynomial ψ(X)− ϕ(Y ) has Schinzel’s property.

So in case of rational curves we have reduced the problem to the study of bivariatepolynomials with separated variables (that is ψ(X)− ϕ(Y )) with Schinzel’s property.

¿From the example of the introduction we can take rational functions r(t), s(t) ∈ Q(t)and define the irreducible minimal polynomial f(X,Y ) such that f(r(t), r(s(t))) = 0;this polynomial has the SP, and the set S is r(t)|t ∈ Q (or also r(t)|t ∈ Z). The mapt 7→ (r(t), r(s(t))) is a minimal parametrization of the curve Cf = (x, y) ∈ A2|f(x, y) =0 which, by construction, is a rational curve.

Lemma 4.1.5 If (ϕ,ψ) and (ϕ1, ψ1) are minimal parametrizations of f(X,Y ) = 0 andψ = ϕ h then ψ1 = ϕ1 h1

So this lemma shows that Conjecture 2 is independent of the chosen parametrizationof a rational curve.

4.2 Symmetric case

The following proposition shows the connection between rational symmetric curvesand polynomials with Schinzel’s property.

4.2. SYMMETRIC CASE 61

Proposition 4.2.1 Let f(X,Y ) ∈ Q[X,Y ] be an irreducible polynomial and supposethat the curve C = (x, y) ∈ A2|f(x, y) = 0 is rational. If f is symmetric then f hasSchinzel’s property and there exists a parametrization of the form

f(ϕ(t), ϕ(a(t))) = 0

Conversely if there exists such a parametrization then f has Schinzel’s property withS = ϕ(t)|t ∈ Q.

Proof : Denote Θ the automorphism of A2 such that Θ(x, y) = (y, x); note that itis an involution: Θ Θ = Id.

Let Φ : A1(Q) → Cf be a parametrization of the curve C, with Φ(t) = (ϕ(t), ψ(t)),ϕ,ψ ∈ Q(t) and inverse λ(x, y), λ ∈ Q(x, y). Since by hypothesis Θ(C) = C, thenΦΘ + Θ Φ is another parametrization of C with inverse λΘ + λ Θ.

Let P = Φ(t) = (ϕ(t), ψ(t)) be a point of the curve C; since

PΘ+ Θ(P ) = ΦΘ(t) = (ψ(t), ϕ(t))

is another point of C and Φ is a parametrization, then there exists t ∈ Q such thatΦ(t) = PΘ, except for a finite number of cases. But

t = λ(PΘ) = λ ΦΘ(t) = λΘ Φ(t) + a(t)

where a(t) ∈ Q(t).Hence Φ(a(t)) = Θ(Φ(t)), so ψ(t) = ϕ(a(t)) and Φ(t) = (ϕ(t), ϕ(a(t))); the set S is

equal to ϕ(t)|t ∈ Q.If the curve C has a parametrization defined over Q (see chapter 4) then S ⊂ Q

otherwise it is contained in the number field K generated over Q by the coefficients ofϕ(t) and a(t).

The following corollary is an immediate consequence.

Corollary 4.2.2 In the same hypothesis and notation of the previous proposition, a(t)has the following expression:

a(t) =αt+ β

γt− α

with α, β, γ ∈ Q e α2 + βγ 6= 0.

Proof : Note that

a a(t) = (λ ΦΘ) (λ ΦΘ) = λ (ΦΘ λΘ) Φ = λ Id Φ = Id(t) = t

so a is an involution: it is a rational function of degree one and it can be easilyproved that a has the desired expression.


Corollary 4.2.3 Let f(X,Y ) ∈ Q[X,Y ] be an irreducible polynomial and suppose thatthe curve C = (x, y) ∈ A2|f(x, y) = 0 is rational.

Then f is symmetric if and only if there exist ϕ, a ∈ Q(t) with deg a = 1 and a2 = Idsuch that (ϕ(t), ϕ(a(t)) is a parametrization of C.

Proof : The ’only if’ part has already been proved in the previous proposition.

The ’if’ part follows from the next two lemmas.

Lemma 4.2.4 Let f(X,Y ) ∈ Q[X,Y ] be an irreducible polynomial and let

C = (x, y) ∈ A2|f(x, y) = 0

be the associated curve in A2(Q). Then f is symmetric if and only if there exists aninfinite subset A ⊂ C such that for every (x, y) ∈ A we have (y, x) ∈ C.

Proof : This lemma follows immediately from proposition 3.1.2: it is sufficientto consider the irreducible curves C and C ′ = (x, y) ∈ A2|g(x, y) = 0, where thepolynomial g(X,Y ) + f(Y,X) is irreducible too. Since by hypothesis C ∩C ′ has infinitepoints it follows that f and g divide each other, so they are equal since they have thesame degree. The lemma is proved.

Lemma 4.2.5 Let r, s ∈ Q(t) be two rational functions and f(X,Y ) ∈ Q[X,Y ] be theminimal polynomial such that f(r(t), r(s(t)) = 0. If s(t) is an involution then f issymmetric.

Proof : If s(t) in an involution then s−1(t) is well defined and is equal to s(t). So if weset τ + s(t) we have

f(r(s(t)), r(t)) = f(r(τ), r(s−1(τ))) = f(r(τ), r(s(τ))) = 0

So for the previous lemma f is symmetric.

Since the rational function s(t) has degree one it determines a birationality of A1

with itself.

Lemma 4.2.6 Let k be a field and f(X,Y ) ∈ k[X,Y ] be the minimal polynomial suchthat f(r(t), r(s(t))) = 0, for r, s ∈ k(t). If f is symmetric then deg s(t) = 1.

Proof : Since f is symmetric then degX(f) = degY (f). The statement follows fromlemma 1.2.1.

4.3. KUBOTA’S RESULTS 63

4.3 Kubota’s results

The only known result in this direction so far are the followings theorems by Kubota(see his articles [21], [22] and [23]).

The main contribution of Kubota in [21] was to show that the following conjectureof Narkiewicz was false (see [33]): let f, g two polynomials in Q[T ] and S ⊂ Q infinitesuch that f(S) = g(S); then deg(f) = deg(g).

Theorem 4.3.1 (Kubota) Let f and g be two polynomials in Q[T ], with deg(g) >deg(f), and suppose that there exists an infinite subset S of Q such that g(S) ⊂ f(S). Ifevery component of the curve g(X)− f(Y ) = 0 containing an infinity of points of S ×Shas a polynomial parametrization, then g = f h for some h ∈ Q[T ].

This theorem is based on the following other one (see [23]):

Theorem 4.3.2 (Kubota) Let f(X) − g(Y ) ∈ Q[X,Y ], where deg(f) 6= deg(g); thenevery component of the curve (x, y)|f(x) = g(y) which admits a polynomial parametriza-tion is of the form f1(X)− g1(Y ), for some polynomials f1, g1 ∈ Q[T ].

Theorem 4.3.3 (Kubota) Let f and g be two polynomials in Q[T ], with n = deg(f)and m = deg(g); suppose also that there exists an infinite sequence S = snn∈N ⊂ Qsuch that g(sn) = f(sn+1) for all n ∈ N. Then m ≥ n.

The following theorem deals with the case of equal degree. A (f, g)− cycle is a finiteset A = aii=0,...,n such that g(ai) = f(ai+1) for all i = 0, . . . , n− 1 and g(an) = f(a0).Note that a (f, g)− cycle is also a (g, f)− cycle.

Theorem 4.3.4 (Kubota) Let f and g be two polynomials in Q[T ], with deg(f) =deg(g); suppose also that there exists an infinite subset S of Q such that f(S) = g(S)with #(f, g)− cycles <∞. Then there exists a linear polynomial h(t) ∈ Q[t] such thatg = f h.

Chapter 5

Parametrization of integer-valued

polynomials

5.1 Introduction

The set of integer-valued polynomials is defined as

Int(Z) = f(X) ∈ Q[X] | f(Z) ⊂ Z

(see for example [6]). It is a Z-module which contains the ring of polynomials overZ. It also contains the binomial polynomials

(X

n

)+X(X − 1) . . . (X − (n− 1))

n!

where n ∈ N.The following theorem, due to Polya, holds

Theorem 5.1.1 The set Int(Z) is a free Z-module. The set of the binomial polynomials(Xn

)| n ∈ N is a basis of Int(Z).

For a proof see [32] or [6].

More explicitly if f ∈ Int(Z) has degree n then we have

f(X) = f(0)

(X

0

)+ ∆f(0)

(X

1

)+ . . .+ ∆nf(0)

(X

n

)

where ∆g(X) = g(X)− g(X + 1) and this expression for f as Z-linear combinationof(Xi

)is unique.

Given a polynomial f ∈ Int(Z) we want to parametrize the set of integer values of fover Z (that is the set f(Z)) with a polynomial with integer coefficients in one or severalvariables. We give the following definition.

65

66 CHAPTER 5. PARAMETRIZATION OF INTEGER-VALUED POLYNOMIALS

Definition 5.1.2 Let f(X) be an integer-valued polynomial.We say that f(Z) is Z-parametrizable if there exists m ∈ N and a polynomialg ∈ Z[X] = Z[X1, . . . , Xm] such that f(Z) = g(Zm).

As an example consider the integer-valued polynomial f(X) = X(X − 1)/2. Wedefine the following polynomials with integer coefficients

g1(X) = f(2X)

g2(X) = f(2X + 1)

Since Z = P ∪D, where P is the set of even integers and D is the set of odd integers,it follows that f(Z) = f(P ) ∪ f(D) = g1(Z) ∪ g2(Z). But f(X) = f(1−X) so we havethat g2(X) = f(−2X) = g1(−X) so g1 and g2 have the same values over the integersand f(Z) can be parametrized by a single polynomial with integer coefficients in onevariable.

It is crucial that f has a symmetry axis of this kind, that is f(h(X)) = f(X), whereh(X) = 1−X, and h swaps P with D; this implies that f(P ) = f(D).

Here we state the main theorem of this chapter.

Theorem 5.1.3 Let f ∈ Int(Z) be of the form f(X) = F (X)/N where F ∈ Z[X] andN ∈ N is minimal.

If N is divisible by a prime number different from 2 then f(Z) is not Z-parametrizable.If N = 2n, where n ∈ N, and f(Z) is Z-parametrizable then there exists β ∈ Q which

is the ratio of two odd integers such that f(X) = f(−X + β).The set f(Z) is equal to g(Z) for some g ∈ Z[X] if and only if f(X) is an integer

coefficient polynomial or it belongs to Z[X(b−X)/2] for some odd integer b.

The content of a polynomial f ∈ Z[X] is defined as the greatest common divisorof the coefficients and it is denoted with cont(f); the hypothesis on N means that(N, cont(F )) = 1. The ring Z(2) is the localization of Z at the prime ideal (2) ⊂ Z; so βis an element of Z∗

(2), the group of invertible elements of Z(2).

Observe also that if f(X) = F (X)/N with N ∈ N and F ∈ Z[X] such that(N, cont(F )) = 1 then f(X) is an integer-valued polynomial if and only if F (Z) ⊂ NZ.

In a paper of Frisch (see [15]) the author deals with parametrization of subsets ofZk with integer coefficient polynomials and integer-valued polynomials. In an anotherarticle of the same author and Vaserstein (see [16]) it is showed that the subset of Z3

of pythagorean triples is parametrizable by a single triple of integer-valued polynomialsin four variables, but it cannot be parametrized by a single triple of integer coefficientspolynomials in any number of variables. Our theorem shows that there exist subsets ofZ parametrizable by an integer-valued polynomial which are not parametrizable by aninteger coefficient polynomial in any number of variables.

The theorem 5.1.3 will be proved in several steps.

5.2. LINEAR FACTORS OF BIVARIATE SEPARATED POLYNOMIALS 67

We need the following theorem by Siegel (see [39] or [25]):

Theorem 5.1.4 (Siegel) Let f ∈ Q[X,Y ] be an irreducible polynomial such thatdegY (f) ≥ 2. Let Ω be the set

Ω + n ∈ Z | ∃ q ∈ Q s.t. f(n, q) = 0

If B ∈ R≥0 then#(Ω ∩ [−B,B]) = O(B1/2)

In particular the set Ω has zero density, that is

limB→+∞

#(Ω ∩ [−B,B])

#(Z ∩ [−B,B])= 0

5.2 Linear factors of bivariate separated polynomials

Given a non-constant polynomial f ∈ Z[X], we define the polynomial in two variables

Ff (X,Y ) +f(X)− f(Y )

X − Y

We remark that f(X)− f(Y ) as a polynomial in the variable Y over the ring Q[X]is separable (for instance by the derivative criterion). This implies that it has distinctirreducible factors in Q[X,Y ].

Proposition 5.2.1 Let f ∈ Z[X]. If for all n ∈ Z there exists q ∈ Q such thatFf (n, q) = 0, then there exists β ∈ Q such that f(X) = f(−X + β).

Proof : Let Ff (X,Y ) =∏i=1,...,s gi(X,Y ) be the factorization of Ff in Q[X,Y ]

and let di be the Y -degree of the factor gi ∈ Q[X,Y ] for i = 1, . . . , s. For each positiveconstant B ∈ R we define the sets

Zi,B + n ∈ Z| |n| ≤ B , ∃ q ∈ Q s.t. gi(n, q) = 0 (5.1)

For those indexes i such that di ≥ 2, each of the sets Zi,B has cardinalityO(B1/2) (thisfollows from Siegel’s theorem 5.1.4). Since by hypothesis Z∩ [−B,B] =

⋃i=1,...,s Zi,B, it

follows that there exists j ∈ 1, . . . , s such that dj = 1, which means that gj(X,Y ) =Y −Qj(X) where Qj ∈ Q[X].

Since Ff (X,Y ) divides f(X)−f(Y ) it follows that f(X) = f(Qj(X)) so deg(Qj) = 1,that is Qj(X) = αX + β with α, β ∈ Q. From the equality f(X) = f(αX + β) wededuce that α = ±1. If α = 1 then it follows immediately that β = 0 (a non constantpolynomial cannot be periodic) but this possibility has to be excluded since X − Y isan irreducible factor of f(X) − f(Y ) that has been removed in Ff (X,Y ). As we haveremarked f(X)− f(Y ) has distinct irreducible factors.


So we have that f(X) = f(−X + β). Observe that the corresponding sets Zi,B areequal to Z ∩ [−B,B].

We easily get the same conclusion if there exists a constant M ∈ R>0 such that forall n ∈ Z, |n| > M , there exists q ∈ Q such that Ff (n, q) = 0.

Corollary 5.2.2 Let f ∈ Q[X]. Then f(X) − f(Y ) ∈ Q[X,Y ] has at most two linearfactors.

Proof : Observe that X − Y divides f(X)− f(Y ) for all f ∈ Q[X].

If α ∈ R is chosen out of a bounded set, the set

Γα = γ ∈ Q|f(γ) = f(α)

has at most two elements, according to the parity of deg(f), since f is definitelystrictly increasing-decreasing. So for |α| ≫ 0 the set Γα has cardinality two if and onlyif deg(f) ≡ 0 (mod 2) and there exists β ∈ Q such that f(X) = f(−X + β) (see theremark after the previous proposition).

If f(X) − f(Y ) =∏gi(X,Y ) is the factorization then f(X) − f(α) =

∏gi(X,α),

where α ∈ R is chosen out of a bounded set. If f(X)− f(Y ) had more than two linearfactors then f(X)− f(α) would have more than two roots which is impossible.

We give also another direct proof of the last fact.

If f(X) = f(−X+β1) = f(−X+β2), where β1 6= β2, then if we set T + −X+β1 wehave that f(T ) = f(T −β1 +β2), which is true if and only if −β1 +β2 = 0 (a polynomialcannot be periodic), which leads to a contradiction.

5.3 Preliminary results

Definition 5.3.1 Let K be a number field and f ∈ K[X] = K[X1, . . . , Xm] of the form

f(X) =∑

i

aiXi

where i = (i1, . . . , im) ∈ Nm and Xi = Xi11 . . . Xim

m .

Let v be a valuation of the field K and | |v the associated norm. The Gauss normof f is defined as (see [34], pg.119):

‖f‖v + maxi|ai|v

5.3. PRELIMINARY RESULTS 69

The Gauss norm coincides with | |v over K. If v is a non-archimedian valuation theGauss norm is multiplicative on K[X] by Gauss Lemma, since we have the equality

‖f‖v = |cont(f)|v

Moreover if x ∈ Omv , where Ov ⊂ K is the valuation ring of v, then it follows that

|f(x)|v ≤ ‖f‖v

Lemma 5.3.2 Let p ∈ N be a prime, let f ∈ Z[X]−pZ[X], let g ∈ Z[X ] = Z[X1, . . . , Xm]and let Q ∈ Q[X] be such that

f(Q(X)) = pg(X)

Then there are two possibilities: either

Q(X) =a0 + pR(X)

D

where a0 ∈ Z, R ∈ Z[X], R(0) = 0 and D ∈ N is such that p ∤ D, or there existalgebraic numbers ξ, π in the splitting field K of f over Q, a non-archimedian valuationv of K above p with valuation ring Ov such that v(π) = 1 and ξ 6∈ Ov, and a polynomialR(X) ∈ Ov[X] such that

Q(X) = ξ(1 + πR(X)).

Proof : Let K be the splitting field over Q of the polynomial f(X) and let v be avaluation of K over the valuation vp of Q with valuation ring Ov ⊂ K and uniformizerπ (that is v(π) = 1). In K[X] the polynomial f factorizes in the following way

f(X) = c0∏

i

(αiX − βi)

where c0 ∈ K , αi, βi ∈ Ov such that (αi, βi) = 1 (remember that Ov is integrallyclosed in K since it is a valuation ring). Note that c0 ∈ Z − pZ since the product ofprimitive polynomials is a primitive polynomial in Ov[X] by Gauss Lemma (Ov is aUFD and K is its quotient field); so the content of f is c0, and c0 is an integer becausef ∈ Z[X]. Since by hypothesis f ∈ Z[X]− pZ[X] then p ∤ c0. Then

f Q(X) = c0∏

i

(αiQ(X)− βi)

and applying the Gauss norm relative to the valuation v

‖f Q(X)‖v =∏

i

‖αiQ(X)− βi‖v = ‖pg(X)‖v < 1


So there is an index j such that ‖αjQ(X)− βj‖v < 1 which implies

‖Q(X)− ξj‖v < |αj |−1v (5.1)

where ξj = βj/αj ∈ K is the corresponding root of f .

Suppose Q(X) =∑

i≥0 aiXi where ai ∈ Q.

First we consider the case π ∤ αj , which implies |αj |v = 1 and ‖Q(X)− ξj‖v < 1.

We have |ai|v = |ai|p < 1 for each i > 0 and |a0− ξj |v < 1. ¿From the first inequalitywe deduce that all the coefficients of Q (except the constant term) are elements of pZ(p),where Z(p) is the localization of Z at the prime ideal (p).

Let us focus on the second inequality. If a0 = a/b, where a, b ∈ Z are coprimeintegers, we have

∣∣∣∣a

b−βjαj

∣∣∣∣v

=

∣∣∣∣αja− bβj

bαj

∣∣∣∣v

=|αja− bβj |v|b|v

< 1

since |αj |v = 1.

Suppose that p|b or equivalently |b|v = |b|p < 1. Then we have |αja−bβj |v < |b|p < 1:this is impossible since a is coprime with b which implies |αja|v = |αj |v|a|p = 1. Hencep ∤ b and so the constant term of Q belongs to Z(p).

So in this case the polynomial Q has the following form

Q(X) =a+ pR(X)

D

where R(X) ∈ Z[X] with R(0) = 0 and D ∈ Z with p ∤ D, as we stated.

Suppose now that π|αj which implies π ∤ βj , since αj and βj are coprime. Hence|βj |v = 1 and ξj = βj/αj 6∈ Ov, which means |ξj |v > 1.

¿From equation (5.1) we have

‖Q− ξj‖v < |ξj |v.

We deduce that ai = ξjπniui for i > 0 and a0 = ξj(1 + πn0u0) where ni > 0 and

ui ∈ O∗v . So there exists a polynomial R ∈ Ov[X] such that

Q(X) = ξj(1 + πR(X))

as we want to prove.

Observe that in the second case of the lemma for all x ∈ Zm we have that

|Q(x)|p = |ξj |v|1 + πR(x)|v = |ξj |v = k > 1

5.3. PRELIMINARY RESULTS 71

since 1+πR(x) ∈ O∗v , where k is a constant independent from the polynomial Q and

the integer vector x. If Q(X) = QD(X)/D, where QD ∈ Z[X] and D ∈ Z such that(D, cont(QD)) = 1, this implies that |QD(x)|p ≤ 1 is constant for all x ∈ Zm and p|D.

This condition is weaker than the second one contained in the lemma: take forexample the polynomial (X2 +1)/2 which is 2-adically constant but it is not of the formof the second case of the lemma.

Lemma 5.3.3 Let H ∈ Q[Y ] and let T be the set

T + n ∈ Z| ∃ y ∈ Zm s.t. n = H(y)

Let v be a non-archimedean absolute value of Q which extends a p-adic absolutevalue | |p of Q.

If T is dense in Q for the topology induced by | |p, then for all γ ∈ Q we have theinequality

‖X − γ‖v ≤ ‖H − γ‖v

Proof : Let γ be an algebraic number and let n ∈ T . Then we have n − γ =(H − γ)(y

n) and so

|n− γ|v ≤ ‖H − γ‖v

since yn∈ Zm.

We define rγ + ‖H − γ‖v.

If |γ|v > 1 then |γ|v = |n− γ|v ≤ rγ .

If |γ|v ≤ 1 then rγ ≥ 1 otherwise we have |n − γ|v < 1 for every n ∈ T . But thismeans that n is constant modulo the valuation v: this is absurd because T ⊂ Q is densefor | |p. In fact let n ∈ T be fixed, then for all n ∈ T we have

|n− n|p = |n− n|v = |n− γ + γ − n|v ≤ max|n− γ|v, |γ − n|v < 1

so T would be contained in a single residue class modulo p.

So we have rγ ≥ max1, |γ|v, which implies that

‖H − γ‖v = rγ ≥ ‖X − γ‖v

This proves the lemma.


5.4 Main results

The following proposition generalizes the example shown in the introduction.

Proposition 5.4.1 Let f ∈ Int(Z) be of the form f(X) = F (X)/2 whereF ∈ Z[X] − 2Z[X]. If there exists an odd integer b such that f(X) = f(−X + b) thenf(Z) is Z-parametrizable.

Proof : Observe that f ∈ Int(Z) is equivalent to F (Z) ⊂ 2Z. In particular 2 dividesa0, the constant term of F (X).

We define the following two polynomials

g1(X) + f(2X) , g2(X) + f(2X + 1)

These polynomials have integer coefficients, in fact if f(X) = (∑

i≥1 aiXi + a0)/2

where ai ∈ Z for i = 0, . . . , n, then

g1(X) = (∑

i≥1

ai2iXi + a0)/2

and it is obviously a polynomial in Z[X]. For the second polynomial we have

2g2(X) =∑

i≥1

ai(2X + 1)i + a0 ≡∑

i≥1

ai + a0 (mod 2Z[X]) = f2(1) ≡ 0 (mod 2Z[X])

which means that g2 ∈ Z[X].

Since Z = P ∪D, where P is the set of even integers and D the set of odd integers,then f(Z) = f(P ) ∪ f(D) = g1(Z) ∪ g2(Z). Now we want to show that g1(Z) = g2(Z).The following equality holds

g2(X) = f(−2X − 1 + b)

Since b− 1 ≡ 0 (mod 2) then g2 has the same image over Z of g1 (if n ∈ Z then form = n+ (1− b)/2 ∈ Z we have g1(n) = g2(m)).

The fact is that the map h : X 7→ −X + b swaps the odd integers with even integersand moreover f(h(X)) = f(X), as we have already remarked.

Proposition 5.4.2 Let q ∈ N let f ∈ Int(Z) be of the form f(X) = F (X)/q whereF ∈ Z[X] and (cont(F ), q) = 1.

If q = 2a, where a ∈ N−0 and f(Z) is Z-parametrizable, then there exists β ∈ Z∗(2)

such that f(X) = f(−X + β).

If q is a prime different from 2 then f(Z) is not Z-parametrizable.

5.4. MAIN RESULTS 73

Proof : Suppose that there exists g ∈ Z[X] such that f(Z) = g(Zm).

Since g(Zm) ⊂ f(Z) then by Hilbert’s irreducibility theorem (see [39]) there existsQ ∈ Q[X] such that f(Q(X)) = g(X). Therefore we can write

F (Q(X)) = pg(X) (5.2)

where p is equal to 2 if q = 2a and p = q if q is a prime different from 2.

From the inclusion f(Z) ⊂ g(Zm) we have that for every n ∈ Z there exists xn ∈ Zm

such that f(n) = g(xn); from this fact and equation (5.2) we deduce

F (n) = F (Q(xn)) (5.3)

Therefore for all n ∈ Z the couple (n,Q(xn)) ∈ Z × (Z/D), where D ∈ Z is thedenominator of Q, is a point of the plane curve

Cf = (x, y) ∈ A2|F (x) = F (y)

Let

F (X)− F (Y ) =∏

i∈I

gi(X,Y )

be the factorization of F (X) − F (Y ) over Q[X,Y ] (remember that there is at leastone linear factor and at most two of them). From (5.3) we have

Z =⋃

i∈I

n ∈ Z | (n,Q(xn)) ∈ Ci

where Ci is the plane curve (x, y) ∈ A2|gi(x, y) = 0. For i ∈ I we define the sets

Ti + n ∈ Z | (n,Q(xn)) ∈ Ci

The sets Ti, for i ∈ I, cover Z. For every B ∈ R, B ≥ 0, and i ∈ I we also define thesets Ti,B + Ti ∩ [−B,B]; they cover ZB, where ZB + Z ∩ [−B,B].

If I ′ is the subset of I of those indexes i such that the Y -degree of gi(X,Y ) is greateror equal than 2, then by a theorem of Siegel (see [39]) we have

#(⋃

i∈I′

Zi,B) = O(B1/2)

where B varies over the real positive numbers and Zi,B’s are the sets defined in(5.1). Since for every i ∈ I we have Ti,B ⊂ Zi,B then #(Ti,B) = O(B1/2) for i ∈ I ′. Inparticular

⋃i∈I′ Ti,B is a set of integers of density zero.

So we focus our attention on the remaining sets Ti for i ∈ I − I ′. They correspondto linear factors of the polynomial F (X)− F (Y ).

Equation (5.2) implies that we can apply lemma 5.3.2.


Suppose that we are in the first case of lemma 5.3.2: Q(X) = a0+pR(X)D , where a0 ∈ Z,

R ∈ Z[X], R(0) = 0 and D ∈ N such that p ∤ D.Let us say that for i = 0 the corresponding factor g0(X,Y ) is X − Y . Then for all

n ∈ T0 we have

n = Q(xn) =a0 + pR(xn)

D

which implies

Dn = a0 + pR(xn) (5.4)

and considering this last equation modulo p we obtain

n ≡ a0D′ (mod p)

where D′ ∈ Z such that DD′ ≡ 1 (mod p).So we have obtained that

T0 ⊂ n ∈ Z| n ≡ a0D′ (mod p) (5.5)

that is T0 is contained in a single residue class modulo p. Therefore #(T0,B) ≤ B/p.As B goes to infinity we have that T0,B ∪

⋃i∈I′ Ti,B does not cover ZB (because

the function B − B1/2 − B/p is definitely positive). By proposition 5.2.1 it followsthat there exists β ∈ Q such that F (X) = F (−X + β), that is I = I ′ ∪ 0 ∪ i0and gi0(X,Y ) = Y + X − β. From corollary 5.2.2 there are no other linear factors ofF (X)− F (Y ) in Q[X,Y ]. We suppose that i0 = 1.

For all n ∈ T1 we have

n = −Q(xn) + β = −a0 + pR(xn)

D+ β

and

Dn = −(a0 + pR(xn)) +Dβ (5.6)

which in particular implies that Dβ = D′′ ∈ Z and β ∈ Z(p), since p ∤ D. Consideringas before this equation modulo p we obtain

n ≡ −a0D′ +D′′D′ (mod p)

hence

T1 ⊂ n ∈ Z| n ≡ −a0D′ +D′′D′ (mod p) (5.7)

and T1 is contained in a single residue class modulo p.If p = 2 then D′′D′ 6≡ 0 (mod 2), otherwise by (5.5) and (5.7) T0 and T1 would becontained in the same residue class, so there would be a residue class not covered by⋃i∈I Ti. Hence if p = 2 we have that β ∈ Z∗

(2).


For each B ≥ 0 we have

0 = #ZB− #(⋃i∈I Ti,B) ≥

#ZB− #(⋃i∈I′ Ti,B) −#(T0,B)−#(Ti0,B) =

B− B1/2 −B/p−B/p =

p− 2

pB −B1/2.

If p is a prime different from 2 we have a contradiction: the sets Ti,B for i ∈ I doesnot cover ZB as B goes to infinity.

In the second case of lemma 5.3.2 we have that for every x ∈ Zm the value |Q(x)|pis a constant greater then one. Then the set T0,B is empty since the equation

n = Q(xn)

has no solution: |n|p ≤ 1 for every n ∈ Z but |Q(xn)|p > 1.

As before by proposition 5.2.1 there exists β ∈ Q such that f(X) = f(−X + β).Therefore Z = Ti0∪(

⋃i∈I′ Ti) and Ti0 contains an Hilbert set, the complement of

⋃i∈I′ Zi,

which is dense in Q for each p-adic absolute value | |p (see Corollary 2.5 of chapter 9 of[25]). Hence we can apply lemma 5.3.3 to the polynomial −Q(X) + β and the set Ti0 :for all γ ∈ Q we have that

‖X − γ‖v ≤ ‖ −Q+ β − γ‖v (5.8)

where v is a valuation in Q which extends the p-adic valuation of Q.

Let

F (X) = c0∏

(X − ξi)

be the factorization of F (X) in Q and fix a valuation v of the splitting field of f over Qover the p-adic valuation of Q.

We have the following relations (here we use inequality (5.8)):

‖F‖v = |c0|v∏‖X − ξi‖v ≤

|c0|v∏‖ −Q+ β − ξi‖v =

‖c0∏

(−Q+ β − ξi)‖v =

‖F (−Q+ β)‖v =

‖F (Q)‖v =

‖pg‖v < 1

but this is a contradiction since F ∈ Z[X]− pZ[X].


Therefore in the second case of lemma 5.3.2 the sets Ti,B for i ∈ I do not cover ZBas B goes to infinity (even in the case p = 2).

If f ∈ Int(Z), f(X) = F (X)/2a, is such that f(Z) is Z-parametrizable, then the firstcase of lemma 5.3.2 holds: in equation (5.2) the polynomial Q(X) belongs to Z(2)[X].

Moreover, if Q(X) = a0+2R(X)D such that a0 ≡ 1 (mod 2) we have that

g(X) = f(Q(X)) = f(−Q(X) + β) = f

(2R(X) + a0 + a′

D

)

since f(X) = f(−X + β), with β ∈ Z∗(2), and Dβ ∈ Z. Observe that a′ ≡ 1 (mod 2).

Hence in any case we may assume that Q(X) ∈ 2Z(2)[X], that is Q(X) = 2R(X)/D.Moreover we have

Z = T0 ∪ T1 ∪ T (5.9)

where T0 ⊂ 2Z, T1 ⊂ 2Z + 1 and T =⋃i∈I′ Ti has zero density.

If n ∈ T0 then n = 2k for some k ∈ Z; by (5.4) we have for all k ∈ Z except for a setof density zero that

Dk = R(x) (5.10)

for some x ∈ Zm.If n ∈ T1 then n = 2k+ 1 for some k ∈ Z; by (5.6) we have for all k ∈ Z except for a

set of density zero that

Dk = −R(x) + (Dβ −D)/2 (5.11)

for some x ∈ Zm.We set α + (Dβ −D)/2 ∈ Z and we remark that α 6≡ 0 (mod D).

Corollary 5.4.3 Let f ∈ Int(Z) be of the form f(X) = F (X)/N where N ∈ N and letF ∈ Z[X] be such that (N, cont(F )) = 1. If there exists a prime p different from 2 suchthat p|N then f(Z) is not Z-parametrizable.

Proof : Suppose that f(Z) is Z-parametrizable and let p be a prime factor of N ,different from 2. Since Int(Z) is a Z-module then g(X) = N

p f(X) ∈ Int(Z) and g(Z) isZ-parametrizable, too. But this is in contradiction with proposition 5.4.2.


5.4.1 Case f(Z) = g(Z), with g ∈ Z[X]

In this section we characterize integer valued polynomials f(X) such that f(Z) isparametrizable with an integer coefficient polynomial g(X) in one variable.

If f ∈ Int(Z) is such that f(Z) = g(Z) for some g ∈ Z[X], then there exists β ∈ Z−2Zsuch that f(X) = f(−X + β), as the following corollary shows.

Corollary 5.4.4 Let f ∈ Int(Z) − Z[X] be of the form f(X) = F (X)/2a, where F ∈Z[X]− 2Z[X] and a ∈ N− 0.

If there exists β ∈ Z∗(2) − Z (that is: β is the ratio of two odd integers) such that

f(X) = f(−X + β) then f(Z) is not Z-parametrizable with an integer coefficient poly-nomial in one variable.

If f(Z) = g(Z) for some g ∈ Z[X] then there exists an odd integer b such thatf(X) = f(−X + b) and g′(X) + f(2X) ∈ Z[X] is such that f(Z) = g′(Z).

Proof : Suppose that there exists β ∈ Z∗(2) − Z such that f(X) = f(−X + β).

We remark that

f(X)− f(Y ) = (X − Y )(X + Y − β)∏

i∈I

gi(X,Y ) (5.12)

where for each i ∈ I the polynomial gi ∈ Q[X,Y ] is irreducible and degY (gi) ≥ 2.Suppose also that there exists g ∈ Z[X] such that f(Z) = g(Z). Then by Hilbert’s

Irreducibility theorem there exist α, δ ∈ Q such that

g(X) = f(αX + δ)

Moreover lemma 5.3.2 and proposition 5.4.2 imply that α, δ ∈ Z(2), with α ∈ 2Z(2).Let us write α = α1/α2, with (α1, α2) = 1, and α2 ≡ 1 (mod 2). Without loss ofgenerality we may suppose also that α2, α1 > 0 (if g(X) parametrizes f(Z) then alsog(−X) parametrizes f(Z)).

Since f(Z) ⊂ g(Z) we have that for each n ∈ Z there exist m ∈ Z such that

f(n) = g(m) = f(αm+ δ)

We define the sets of integers

T0 + n ∈ Z| n = αm+ δ for some m ∈ Z

T1 + n ∈ Z| n+ αm+ δ = β for some m ∈ Z

T2 + n ∈ Z|∃ i ∈ I,m ∈ Z s.t. gi(n, αm+ δ) = 0

By (5.12) we have Z = T0 ∪ T1 ∪ T2; by Siegel’s theorem 5.1.4, T2 is a subset of Zof zero density. We saw in the proof of proposition 5.4.2 (see also (5.9)) that T0 and T1

cannot be empty.


For n ∈ T0 there exists m ∈ Z such that n = αm+δ; if we multiply this last equationby α2 we have that α2δ ∈ Z.

For n ∈ T1 there exists m ∈ Z such that n + αm + δ = β. Hence α2β ∈ Z, as wehave already seen in proposition 5.4.2. Since we are assuming that β 6∈ Z and α2 is anodd integer we have that α2 ≥ 3.

Since g(Z) ⊂ f(Z) we have that for each n ∈ Z there exist m ∈ Z such that

g(n) = f(αn+ δ) = f(m)

Like before we define the sets of integers

S0 + n ∈ Z| αn+ δ = m for some m ∈ Z

S1 + n ∈ Z| αn+ δ +m = β for some m ∈ Z

S2 + n ∈ Z| ∃ i ∈ I,m ∈ Z s.t. gi(αn+ δ,m) = 0

By (5.12) we have Z = S0∪S1∪S2. For all i ∈ I we define g′i(X,Y ) + gi(αX+δ, Y ) ∈Q[X,Y ] which is an irreducible polynomial of degree in Y greater or equal to 2; by Siegel’stheorem 5.1.4, S2 is a subset of Z of zero density.

If n ∈ S0 we have αn + δ ∈ Z, hence α1n + δα2 ≡ 0 (mod α2) (remember thatδα2 ∈ Z), so

n ≡ −δα2k (mod α2)

where k ∈ Z, kα1 ≡ 1 (mod α2) (such k exists since α1 and α2 are coprime).

For each n ∈ S1 we have αn+ δ − β ∈ Z so

n ≡ −δα2k + βα2k (mod α2)

(remember that βα2 ∈ Z).

Since α2 ≥ 3 there are residue class modulo α2 which are not covered by S0 or S1:contradiction.

In particular, if f ∈ Int(Z)−Z[X] is such that f(Z) = g(Z) with g ∈ Z[X], then thereexists an odd integer β such that f(X) = f(−X + β). Moreover g(X) = f(αX + δ),where α = 2 and δ ∈ Z. In fact α ∈ Z (otherwise S0, S1, S2 do not cover Z), whichimplies that δ ∈ Z, since for each n ∈ T0 we have that n − αm = δ, for some m ∈ Z.Moreover for each n ∈ T0 we have that n ≡ −δ (mod α) and for each n ∈ T1 we havethat n ≡ β − δ (mod α). If α 6= 2 there are residue classes which are not covered by T0

and T1.We may also assume that δ = 0. In fact in general if f ∈ Q[X] such that f(X) = f(b−X)for some odd integer b = 2m+ 1 then f(2X + δ) ∈ Z[X] for some integer δ if and onlyif f(2X) ∈ Z[X]. If δ ≡ 0 (mod 2) then f(2X + δ) = f(2(X + k)) + g(X) ∈ Z[X]; then


g(X−k) = f(2X) ∈ Z[X]. If δ ≡ 1 (mod 2) then f(2X+δ) = f(2(X+k)+1) + g(X) ∈Z[X]; then g(−X − k+m) = f(−2X + b) ∈ Z[X]. And this implies that f(2X) ∈ Z[X].Finally it is easy to observe that if g(X) = f(2X + δ) parametrizes f(Z) for some δ ∈ Zthen h(X) = f(2X + δ′) ∈ Z[X] parametrizes f(Z) for every δ′ ∈ Z.

Proposition 5.4.5 Let f ∈ Int(Z)− Z[X].Then f(Z) = g(Z) for some polynomial g ∈ Z[X] if and only if there exists an odd

integer b such that f(X) = f(−X + b) and f(2X) ∈ Z[X].

Proof : If f(Z) = g(Z) with g ∈ Z[X], then there exists b ∈ Z − 2Z such thatf(X) = f(−X + b) (proposition 5.4.2 and corollary 5.4.4). We can assume thatg(X) = f(2X) (see corollary 5.4.4).

If there exists an odd integer b such that f(X) = f(−X+ b) and f(2X) ∈ Z[X] thenwe define g(X) + f(2X). We have that Z = (2Z)∪(2Z+1), so f(Z) = f(2Z)∪f(2Z+1).

We have f(h(X)) = f(X), where h(X) = −X + b. Then

f(2Z) = f(h(2Z)) = f(2Z + 1)

So f(Z) = g(Z).

Note that for all f ∈ Int(Z) of the form f(X) = F (X)/2 we have that f(2X) ∈ Z[X].If f(X) = F (X)/2n, n > 1, the condition f(X) = f(b −X), for some integer b ≡ 1

(mod 2) is not sufficient in order for f(Z) to be equal to g(Z), for some g ∈ Z[X].For example take the polynomial f(X) = X(X − 1)(X − 2)(X − 3)/8.

Lemma 5.4.6 Let b ∈ Z. Then we have

f ∈ Z[X] | f(X) = f(−X + b) = Z[X(b−X)]

Proof : The inclusion ⊃ is obvious.Let f ∈ Z[X] be such that f(X) = f(−X+b). We remark that the degree n of f(X)

is an even integer 2m. We prove that f ∈ Z[X(b−X)] by induction on m.Since the map σ : X 7→ −X + b is a homomorphism of the ring Z[X] in itself, we

may suppose that f(0) = f(b) = 0 (we write f(X) = (f(X)− f(0)) + f(0)).If m = 1 we obviously have that f(X) = aX(b − X), for some a ∈ Z. Let now thestatement holds for every polynomial of degree less than 2m and take a polynomialf(X) of degree 2(m+ 1) such that f(X) = f(−X + b). We have that

f(X) = X(b−X)g(X)

where g(X) ∈ Z[X] of degree 2m such that g(X) = g(−X + b), since σ(X(b−X)) =X(b − X). By inductive hypothesis g(X) = P (X(b − X)), for some P ∈ Z[T ]. Thelemma is proved.


Lemma 5.4.7 Let b be an odd integer and a a positive integer. Let f(X) = F (X)/2a,where F ∈ Z[X]− 2Z[X], F (X) = F (−X + b), such that f(2X) ∈ Z[X].Then a ≤ deg(f)/2.

Proof : We remark that the degree of f is an even integer 2m. We prove that a ≤ mby induction on m. Let m = 1, then by lemma 5.4.6

f(X) =F (X)

2a=A1(X(b−X)) +A0

2a

where A0, A1 ∈ Z. By hypothesis on F (X) we have that A1 and A0 are not botheven integer. So

f(2X) =A1(2X(b− 2X)) +A0

2a=

A1

2a−1X(b− 2X) +

A0

2a

This polynomial belongs to Z[X] if and only if a = 1 (note that since A0 ≡ 0(mod 2a) then A1 ≡ 1 (mod 2); moreover b ≡ 1 (mod 2), so b− 2X is primitive).

Let now the statement holds for those polynomial f(X) of degree less or equal to 2mand consider now f(X) = F (X)/2a of degree 2(m+ 1). If a = 1 we are done. Supposenow a > 1.

We have that

f(X) =F (X)

2a=X(b−X)

2

G(X)

2a−1+F (0)

2a

where G ∈ Z[X] of degree 2m such that G(X) = G(b−X). Since f(2X) ∈ Z[X] wehave that F (0)/2a ∈ Z; hence G 6∈ 2Z[X]. We have

f(2X) =2X(b− 2X)

2

G(2X)

2a−1+F (0)

2a= X(b− 2X)

G(2X)

2a−1+F (0)

2a

and this belongs to Z[X] if and only if G(2X)/2a−1 ∈ Z[X] (since X(b − 2X) isprimitive). By inductive hypothesis applied to the polynomial G(X)/2a−1 we have thata− 1 ≤ m. The lemma is proved.

Lemma 5.4.8 Let f ∈ Q[X], f(X) = F (X)/N where N ∈ Z, F ∈ Z[X] such that(N, cont(F )) = 1. Let f(2X) ∈ Z[X]. If p is a prime different from 2 then p ∤ N .

Proof : We remark that f(2X) ∈ Z[X] if and only if F (2X) ∈ NZ[X].Let us write F (X) = anX

n + . . .+ a0, where ai ∈ Z for i = 0, . . . , n.By hypothesis for each prime q which divides N we have that ‖F‖q = max|ai|q = 1.Suppose now that there exists a prime p different from 2 such that p|N .

If G(X) + F (2X) we have ‖G‖p = max|ai2i|p = max|ai|p = 1. Hence G 6∈ qZ[X],

so G 6∈ NZ[X]: contradiction.


Lemma 5.4.9 Let f ∈ Q[X] be such that f(X) = f(−X + b) for some odd integer band f(2X) ∈ Z[X]. Then f ∈ Int(Z).

Proof : We have to prove that f(Z) ⊂ Z. The proof easily follows from the factthat

f(Z) = f(2Z) ∪ f(2Z + 1)

It is clear that f(2Z) ⊂ Z. We have to show that f(2Z + 1) ⊂ Z. But this followsfrom the fact that f(2Z + 1) = f(2Z + 1 + b) = f(2Z).

We summarize the previous three lemmas in the following proposition.

Proposition 5.4.10 Let f ∈ Q[X] be such that f(X) = f(−X+b) for some odd integerb and f(2X) ∈ Z[X]. Then we have that f ∈ Int(Z) and f(X) = F (X)/2a for someinteger a such that 0 ≤ a ≤ deg(f)/2.

Note that if f ∈ Int(Z), f(X) = F (X)/2a such that f(X) = f(b−X) for some integerb it is not true in general that f ∈ Z[X(b−X)/2]. Take for example the polynomial

f(X) =X(X − 1)(X − 2)(X − 3)

8

which satisfies f(X) = f(3−X) but does not belong to Z[X(3−X)/2].

Proposition 5.4.11 Let b be an odd integer. Then we have

f ∈ Int(Z) | f(X) = f(−X + b), f(2X) ∈ Z[X] = Z

[X(b−X)

2

]

Proof : Let us call Sb the set of the left member.

If f ∈ Z[X(b−X)/2] it is clear that f ∈ Sb.Let now f ∈ Sb. Observe that deg(f) = 2m for some m ∈ N.

By lemma 5.4.8 we have that f(X) = F (X)/2a, for some a ∈ N and F ∈ Z[X]. Ifa ≥ 1 we assume that F 6∈ 2Z[X], that is f(X) is written in reduced form.

We prove by induction on m that f ∈ Z[X(b−X)/2].

Let m = 1; if a = 0 we are done, since Z[X(b−X)] ⊂ Z[X(b−X)/2]. Suppose thata ≥ 1; by lemma 5.4.6 we have F (X) = A1X(b−X) +A0 for some A0, A1 ∈ Z. Then

f(X) = A1X(b−X)

2a+A0

2a

Since f(2X) ∈ Z[X] we have that A0 ≡ 0 (mod 2), so A1 ≡ 1 (mod 2) (becauseF 6∈ 2Z[X]). In particular a = 1, so f ∈ Z[X(b−X)/2].


Let now the statement holds for those polynomial in Sb of degree less or equal to 2mand take a polynomial f ∈ Sb of degree 2(m + 1). If a = 0 we are done; otherwise wehave (again by lemma 5.4.6):

f(X) =X(b−X)

2

G(X)

2a−1+A0

2a

where A0 ∈ Z and G ∈ Z[X] of degree 2m, such that G(X) = G(−X + b). As beforeA0 ≡ 0 (mod 2), so G 6∈ 2Z[X]. We set g(X) + G(X)/2a−1.

We have

f(2X) = X(b− 2X)G(2X)

2a−1+A0

2a

and this polynomial belongs to Z[X] if and only if g(2X) = G(2X)/2a−1 ∈ Z[X] (thepolynomial X(b− 2X) is primitive). We summarize the properties of g(X):

• g ∈ Int(Z), by lemma 5.4.9

• g(X) = g(−X + b)

• g(2X) ∈ Z[X]

Therefore we may apply inductive hypothesis to g(X): g(X) = P (X(b−X)/2), forsome P ∈ Z[T ]. Hence f(X) = P ′(X(b−X)/2), where P ′(T ) = TP (T ) + F (0).

The proposition is proved.

The following theorem characterizes integer valued polynomials f(X) such thatf(Z) = g(Z) for some g ∈ Z[X].

Theorem 5.4.12 We have

f ∈ Int(Z)| f(Z) = g(Z) for some g ∈ Z[X] = Z[X]⋃(

⋃

b∈2Z+1

Z

[X(b−X)

2

])

Proof : If f ∈ Z[X(b−X)/2], where b is an odd integer, then f ∈ Int(Z). We alsohave that f(X) = f(−X + b) and f(2X) ∈ Z[X], so by proposition 5.4.5 we are done.

Conversely let f ∈ Int(Z) − Z[X] (the case f ∈ Z[X] is obvious) be such thatf(Z) = g(Z) for some g ∈ Z[X]. We saw in proposition 5.4.5 that there exists an oddinteger b such that f(X) = f(−X + b) and f(2X) ∈ Z[X]. By proposition 5.4.11 wehave f ∈ Z[X(b−X)/2]. The theorem is proved.


5.4.2 General case

In this last section we show that there exist integer valued polynomials f(X) suchthat f(Z) is Z-parametrizable with an integer coefficient polynomial g(X) which hasmore than one variable, but f(Z) 6= g(Z) for every g ∈ Z[X].

Lemma 5.4.13 Let f(X) = pkX(pkX − a)/2, where p is a prime different from 2, k apositive integer and a an odd integer such that (a, p) = 1. Then f ∈ Int(Z) and f(Z) isZ-parametrizable.

Proof : It is easy to verify that f ∈ Int(Z), since a and p are odd.We define the polynomial

Q(X1, X2) +2

pk(pkX1 + γ(X

k(p−1)2 − 1)k)

where γ ∈ Z is such that (−1)kγ ≡ α (mod pk), with α + (a − pk)/2 ∈ Z. Let uswrite α− (−1)kγ = pkA, for some A ∈ Z.

We define g(X1, X2) + f(Q(X1, X2)). It is easy to check that g ∈ Z[X1, X2].We state that f(Z) = g(Z2). We remark that

f(X)− f(Y ) =pk

2(X − Y )(pkX + pkY − a) =

p2k

2(X − Y )(X + Y −

a

pk)

Let n ∈ Z; we claim that there exists m = (m1,m2) ∈ Z2 such that

f(n) = g(m1,m2) = f(Q(m1,m2))

If n = 2h for some h ∈ Z, then we set m = (h, 1); in this case we have that

n = Q(m)

If n = 2h+ 1 for some h ∈ Z, then we set m = (−h+A, 0); in this case we have

n = −Q(m) +a

pk

Conversely, let m = (m1,m2) ∈ Z2. Then there exists n ∈ Z such that

g(m) = f(Q(m)) = f(n)

In fact if m2 6≡ 0 (mod p) then ((mk2)p−1 − 1)k ≡ 0 (mod pk), by Fermat’s little

theorem. Therefore

Q(m) =2

pk(pkm1 + pkM) = 2(m1 +M)

for some M ∈ Z. We choose n = 2(m1 +M) so we have that


Q(m) = n

If m2 ≡ 0 (mod p) then mk(p−1)2 ≡ 0 (mod pk), and so (m

k(p−1)2 − 1)k ≡ (−1)k

(mod pk). Hence

Q(m) =2

pk(pkm1 + (−1)kγ + pkM)

for some M ∈ Z. We choose n = 2(−m1 −M +A) + 1 so we have that

Q(m) = −n+a

pk

The lemma is proved.

If f(X) = pkX(pkX − a)/2 with k ≥ 1, we remark that f(X) = f(−X + a/pk).Hence by corollary 5.4.4 the set f(Z) is not equal to g(Z), for every g ∈ Z[X]. We canalso use theorem 5.4.12: since f(X) does not belong to Z[X(b − X)/2], for every oddinteger b, then f(Z) 6= g(Z), for every g ∈ Z[X].

Corollary 5.4.14 Let p be an odd prime, k a non negative integer and a an odd integer,coprime with p. Then

Z

[pkX(pkX − a)

2

]⊂ f ∈ Int(Z)| f(Z) is Z-parametrizable

Proof : The statement follows easily from the previous lemma and from the factthat if f(Z) is Z-parametrizable and P ∈ Z[X], then P (f(Z)) is Z-parametrizable.

Lemma 5.4.15 Let f(X) = bX(bX − a)/2, where a and b are odd coprime integers,both different from zero.If f(Z) is Z-parametrizable then b = pk, where p is a prime and k ∈ Z, k ≥ 0.

Proof : Note that f ∈ Int(Z) and f(X) = f(−X + β), where β = a/b.Suppose that b = b1b2, where b1, b2 ∈ Z− ±1 such that (b1, b2) = 1. Suppose also

that f(Z) = g(Zm), for some g ∈ Z[X1, . . . , Xm] = Z[X].We saw in proposition 5.4.2 that g(X) = f(2R(X)/D), for some R ∈ Z[X] and

D ∈ Z which is odd and (D, cont(R)) = 1 (see also the remarks after proposition 5.4.2).

Since g(Zm) ⊂ f(Z) we have that for each x ∈ Zm there exists n ∈ Z such that

g(x) = f(Q(x)) = f(n)

where Q(X) = 2R(X)/D.


We have that f(X)− f(Y ) = (b2/2)(X − Y )(X + Y − β). Hence

f(Q(X))− f(Y ) =b2

2(Q(X)− Y )(Q(X) + Y − β)

We define the sets

S0 + x ∈ Zm | Q(x) = m for some m ∈ Z = x ∈ Zm | R(x) = Dk for some k ∈ Z

S1 + x ∈ Zm | Q(x) = −m+ β for some m ∈ Z = x ∈ Zm | R(x) = −Dk + α for some k ∈ Z

where α = (Dβ −D)/2 ∈ Z (see also (5.10) and (5.11)). So Zm = S0 ∪ S1.We recall that since f(Z) ⊂ g(Zm) then Z = T0 ∪ T1, where

T0 = n ∈ Z |n = Q(x) for some x ∈ Zm

T1 = n ∈ Z |n = −Q(x) + β for some x ∈ Zm

We remark that Si 6= ∅ since Ti 6= ∅, for i = 0, 1. Moreover, if i ∈ 0, 1 andx ∈ Si, then x′ ∈ Si for all x′ ≡ x (mod D) (which means x′j ≡ xj (mod D) for allj = 1, . . . ,m).

Since T1 is not empty we saw in (5.6) that Dβ ∈ Z, that is D = bb′ = b1b2b′,

where b′ ∈ Z. Moreover α 6≡ 0 (mod D) (see remarks after proposition 5.4.2) andα = b′(a− b)/2 with (b, (a− b)/2) = 1; we can find two coprime factors d1, d2 of D suchthat D = d1d2 and α 6≡ 0 (mod di) for i = 1, 2.

Let x ∈ S0, that is R(x) ≡ 0 (mod D): then R(x) ≡ 0 (mod d1).So R(x′) ≡ 0 (mod d1) for all x′ ≡ x (mod d1). For such x′’s we have that x′ 6∈ S1,since α 6≡ 0 (mod d1). So x′ ∈ S0 for all x′ ≡ x (mod d1).

Therefore R(x′) ≡ 0 (mod d2) for all x′ ≡ x (mod d1).We state that for all y ∈ Zm we have R(y) ≡ 0 (mod d2). In fact if we consider the

natural projection map

π1 : Zm → (Z/d1Z)m

and the isomorphism (because (d2, d1) = 1)

Φ : (Z/d1Z)m → (Z/d1Z)m

x 7→ d2x

We have

π1(y − x) = Φ(π1((x)) = d2π1(x)

for some x ∈ Zm. Hence


y − x = d2x+ d1x′

for some x′ ∈ Zm. If we set x′ + x+ d1x′, we have that R(y) ≡ R(x′) (mod d2).

But x′ ≡ x (mod d1), so R(y) ≡ 0 (mod d2). Since α 6≡ 0 (mod d2) this would implythat S1 is empty, contradiction.

For example this lemma shows that the polynomial f(X) = 15X(15X − 1)/2 is inInt(Z), f(X) = f(−X + 1/15) but f(Z) is not Z-parametrizable.

From corollary 5.4.14 and the previous lemma we can state the following conjecture.

ConjectureWe have

f ∈ Int(Z)| f(Z) is Z-parametrizable = Z[X]⋃

⋃

a∈2Z+1, p 6=2

(a,p)=1, k≥0

Z

[pkX(pkX − a)

2

]

where p runs over the set of odd primes.

The inclusion ⊃ follows from corollary 5.4.14.

If f ∈ Int(Z)− Z[X] is such that f(Z) = g(Zm) for some g ∈ Z[X], then proposition5.4.2 shows that:

• f(X) = F (X)/2n for some n ≥ 1 and F ∈ Z[X]− 2Z[X]

• there exists β ∈ Z∗(2) such that f(X) = f(−X + β)

5.5 Number field case

We give a conjecture in the number field case.Let K be a number field with ring of integers OK . We define the set

Int(OK) = f ∈ K[X] | f(OK) ⊂ OK

Definition 5.5.1 Let f ∈ K[X] be an integer-valued polynomial, that is f ∈ Int(OK).We say that f(OK) is OK-parametrizable if there exists m ∈ N and a polynomialg ∈ OK [X]K [X1, . . . , Xm] such that f(OK) = g(OmK).

5.5. NUMBER FIELD CASE 87

This lemma is a generalization of lemma 5.3.2.

Lemma 5.5.2 Let f ∈ OK [X], g ∈ OK [X], a ∈ K such that

f(Q(X)) = ag(X)

Let v be a non archimedean valuation of K with uniformizer π and valuation ringOv such that v(a) ≥ 1, ‖f‖v = 1.

Then there are two possibilities: either

Q(X) =a0 + πR(X)

D

where a0 ∈ Ov, D ∈ O∗v, R ∈ Ov[X], R(0) = 0 or there exist algebraic numbers ξ, π′

in the splitting field K ′ of f over K, a valuation v′ of K ′ above v with valuation ringOv′ such that v(π′) = 1 and ξ 6∈ Ov′, and a polynomial R(X) ∈ Ov′ [X] such that

Q(X) = ξ(1 + π′R(X)).

Conjecture:Let f ∈ Int(OK) − OK [X] be of the form f(X) = F (X)/a, where F ∈ OK [X] and

a ∈ OK .Let v a non archimedean valuation of K such that v(a) ≥ 1 and ‖F‖v = 1. Suppose

also that f(OK) is OK-parametrizable.Then N(v) = q = pf (the cardinality of the residue field of v) is less or equal than

the number of linear factors of f(X)− f(Y ) in the ring K[X,Y ].

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